AP 10th Class Maths Chapter 8 Important Questions Introduction to Trigonometry

These AP 10th Class Maths Chapter Wise Important Questions 8th Lesson Introduction to Trigonometry will help students prepare well for the exams.

8th Lesson Introduction to Trigonometry Class 10 Important Questions with Solutions

10th Class Maths Introduction to Trigonometry 1 Mark Important Questions

Question 1.
Find the value of θ in 2 cos 3θ = 1.
Solution:
Given 2 cos 3θ = 1
cos 3θ = \(\frac{1}{2}\) = cos 60°
3θ = 60° (∵ cos 60° = \(\frac{1}{2}\))
θ = \(\frac{60}{3}\) = 20°

Question 2.
Evaluate \(\frac{\sin 20^{\circ}}{\cos 70^{\circ}}\).
Solution:
\(\frac{\sin 20^{\circ}}{\cos 70^{\circ}}\) = \(\frac{\sin 20^{\circ}}{\cos \left(90^{\circ}-20^{\circ}\right)}\) = \(\frac{\sin 20^{\circ}}{\sin 20^{\circ}}\) = 1 (∵ cos (90 – θ) = sin θ)

Question 3.
Prove that (1 – sin²θ) sec²θ = 1.
Solution:
LHS = (1 – sin²θ) sec²θ
= cos²θ.sec²θ (∵ 1 – sin²θ = cos²θ)
= (cos θ.sec θ)² (∵ cos θ.sin θ = 1)
= 1 = RHS

Question 4.
Prove that cos²θ (1 + tan²θ) = 1.
Solution:
LHS = cos²θ (1 + tan²θ) (∵ 1 + tan²θ = sec²θ)
= cos²θ.sec²θ (cos θ × sec θ = 1)
= (cos θ.sec θ)²
= 1² = 1 = RHS

Question 5.
Prove that (1 – cos²A) cosec²A = 1.
Solution:
LHS = (1 – cos²A) cosec²A (∵ 1 – cos²A = sin²A)
= sin² A.cosec²A
= (sin A.cosec A) = 1² = 1 (sin A.cosec A = 1)

Question 6.
Prove that (1 + cot²A) sin²A = 1
Solution:
LHS = (1 + cot²A) sin²A (∵ 1 + cot²A = cosec² A)
= cosec²A.sin²A
= (cosec A.sin A)² = 1² = 1 (∵ cosec A.sin A = 1)

Question 7.
What is the value of sin²θ + \(\frac{1}{1+\tan ^2 \theta}\).
Solution:
sin²θ + \(\frac{1}{1+\tan ^2 \theta}\)
We know 1 + tan²θ = sec²θ
= sin²θ + \(\frac{1}{\sec ^2 \theta}\) [∴ \(\frac{1}{\sec \theta}\) = cosθ]
= sin²θ + cos²θ
= 1
Therefore, sin²θ + \(\frac{1}{1+\tan ^2 \theta}\) = 1

Question 8.
\(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30}\) =
Solution:

Question 9.
If tanθ = \(\frac{5}{2}\) then \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) =
Solution:

Question 10.
If cosA = \(\frac{4}{5}\) then find the value of tan A.
Solution:

cosA = \(\frac{4}{5}\)
BC = \(\sqrt{5^2-4^2}\)
BC = \(\sqrt{25-16}\) = √9 = 3
tanA = \(\frac{3}{4}\).

Question 11.
(1 – cos²A) =
Solution:
(1 – cos²A) = sin²A

Question 12.
If sinθ = \(\frac{3}{4}\), then \(\frac{\left(\sec ^2 \theta-1\right) \cos ^2 \theta}{\sin \theta}\) =
Solution:

Question 13.
If cot θ = \(\frac{1}{\sqrt{3}}\) find the value of sec²θ + cosec²θ.
Solution:
cotθ = \(\frac{1}{\sqrt{3}}\)
cotθ = cot 60°
θ = 60°
sec²θ + cosec²θ = sec²60° + cosec²60°
= (2)² + (\(\frac{2}{\sqrt{3}}\))²
= 4 + \(\frac{4}{3}\) = \(\frac{16}{3}\) = 5\(\frac{1}{3}\)

Question 14.
Given that secθ = √2, find the value of \(\frac{1+\tan \theta}{\sin \theta}\).
Solution:
secθ = √2
cosθ = \(\frac{1}{\sqrt{2}}\) = cos 45° ⇒ θ = 45°
\(\frac{1+\tan \theta}{\sin \theta}\) = \(\frac{1+\tan 45^{\circ}}{\sin 45^{\circ}}\) = \(\frac{1+1}{\frac{1}{\sqrt{2}}}\)
= \(\frac{2}{\frac{1}{\sqrt{2}}}\) = 2√2

Question 15.
If θ is an acute angle and tanθ + cotθ = 2, then find the value of sin³θ + cos³θ.
Solution:
tanθ + cotθ = 2
1 + 1 = 2, θ = 45°
sin³θ + cos³θ = sin³45° + cos³45°
= (\(\frac{1}{\sqrt{2}}\))³ + (\(\frac{1}{\sqrt{2}}\))³
= \(\frac{1}{2 \sqrt{2}}\) + \(\frac{1}{2 \sqrt{2}}\)
= \(\frac{2}{2 \sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)
sin³θ + cos³θ = \(\frac{\sqrt{2}}{2}\)

Question 16.
If acotθ + bcosecθ = p and bcotθ + acosecθ = q, then p² – q² =
Solution:
p² = (acotθ + bcosecθ)²
= a²cot²θ + 2abcotθ cosecθ + b²cosec²θ
q² = (bcotθ + acosecθ)²
= b²cot²θ + 2abcotθ cosecθ
+ a²cosec²θ
p² – q² = a²cot²θ + 2ab cotθ cosecθ + b² cosec²θ
– (b²cot²θ + a² cosec²θ + 2ab cotq cosecθ)
= cot²θ (a² – b²) + cosec²θ (b² – a²)
= cot²θ (a² – b²) + cosec²θ (b² – a²)
= cosec²θ (b² – a) – cot(b² – a²) [∵ cosec²θ – cot²θ = 1 ]
= b² – a² (1)
= b² – a²

Question 17.
The value of (sin²θ + \(\frac{1}{1+\tan ^2 \theta}\)) =
Solution:
sin²θ + \(\frac{1}{1+\tan ^2 \theta}\) = sin²θ + \(\frac{1}{\sec ^2 \theta}\)
sin²θ + cos²θ
= 1

Question 18.
The value of (1 + tan²θ) (1 – sinθ) (1 + sinθ) = ……………
Solution:
(1 + tan²θ) (1 – sinθ) (1 + sinθ)
= sec²θ (1 – sin²θ)
sec²θ × cos²θ = \(\frac{1}{\cos ^2 \theta}\).cos²θ = 1.

Question 19.
If sinA + sin²A = 1, then find the value of the expression (cos²A + cos⁴A).
Solution:
sinA + sin²A = 1
sinA = 1 – sin²A
sinA = cos²A
cos²A + cos⁴A = cos²A + (cos²A)²
= cos²A + sin²A = 1.

Question 20.
8cot²θ – 8cosec²A =
Solution:
8 cot²A – 8 cosec²A
= 8 (cot²A – cosec²A)
= 8 (-1) – (∵ cosec²θ – cot²θ = 1)
= -8

Question 21.
Find the value of (tan² 45° – cos² 60°).
Solution:
tan² 45° – cos² 60°
= 1² – (\(\frac{1}{2}\))² = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)

Question 22.
In a right-angled triangle PQR, ∠Q = 90°. If ∠P = 45°, then find the value of tanP – cos²R.
Solution:

∠R = 180° – 90° – 45°
∠R = 45°
tanP – cos²R = tan 45 – cos² (45°)
= 1 (\(\frac{1}{\sqrt{2}}\))² = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 23.
If tanθ = \(\frac{2}{3}\), then find the value of secθ.
Solution:

tanθ = \(\frac{2}{3}\)
AC = \(\sqrt{2^2+3^2}\)
AC = \(\sqrt{4+9}\) = \(\sqrt{13}\)
secθ = \(\frac{\sqrt{13}}{3}\)

Question 24.
If sinA = \(\frac{1}{2}\), then find the value of secA.
Solution:
sinA = \(\frac{1}{2}\)
sin A = sin 30°
A = 30°
∴ sec 30° = \(\frac{2}{\sqrt{3}}\)

Question 25.
What is the value of
cos 1° . cos 2° cos 3°. cos 4°…… cos 90° ?
Solution:
cos1° cos 2° cos 3°… cos 90°
= cos1° cos 2° cos 3°… 0 = 0.

Question 26.
Find the value of 2 sin² 30° + 3 tan² 60°- cos² 45°.
Solution:
2 sin² 30 + 3 tan² 60° – cos² 45°
= 2(\(\frac{1}{2}\))² + 3(√3)² – (\(\frac{1}{\sqrt{2}}\))²
= 2 × \(\frac{1}{4}\) + 3 × 3 – \(\frac{1}{2}\)
= \(\frac{1}{2}\) + 9 – \(\frac{1}{2}\)
= 9

Question 27.
In, ΔABC, right angled at C, if tan A = \(\frac{8}{7}\), then find the value of cot B.

Solution:
cot B = \(\frac{1}{\tan \mathrm{A}}\)
cot B = \(\frac{8}{7}\)

Question 28.
Write the value of cot² θ – \(\frac{1}{\sin ^2 \theta}\).
Solution:
cot² θ – \(\frac{1}{\sin ^2 \theta}\)
= cot² θ – cosec² θ
= -(cosec² θ – cot² θ) = -1.

Question 29.
sinA = 8/17 then the value of cosA cosecA = ………………. .
Solution:

AC = \(\sqrt{17^2-8^2}\)
AC = \(\sqrt{289 – 64}\)
AC = \(\sqrt{225}\) = 15
cosA . cosecA = \(\frac{18}{17}\) × \(\frac{17}{15}\) = \(\frac{8}{15}\)

Question 30.
If cosec A + cotA = \(\frac{1}{3}\) then cosecA – cotA = ……………
Solution:
cosec A + cot A = 1/3
cosec²A – cot²A = 1
(cosec A + cotA) (cosec A – cot A) = 1
1/3 (cosec A – cot A) = 1
cosec A – cot A = 3

Question 31.
sin 2A = \(\frac{1}{2}\) tan² 45° then find the value of A.
Solution:
sin2A = \(\frac{1}{2}\) tan²45°
sin2A = \(\frac{1}{2}\)(1)²
sin2A = \(\frac{1}{2}\)
sin2A = sin30°
2A = 30°
A = 15°.

Question 32.
If sinθ = 0.8 then find the value of cosθ.
Solution:

sinθ = 0.8 = \(\frac{8}{10}\) = \(\frac{4}{5}\)
x = \(\sqrt{5^2-4^2}\)
x = \(\sqrt{25 – 16}\) = √9 = 3
∴ cosθ = \(\frac{3}{5}\)

Question 33.
How much value of the angle ‘θ’ in the figure ?
Answer:
30°

Question 34.
Observe the following:
(I) sin²20° + sin²70° = 1
(II) log₂ (sin 90° = 1
Which one is CORRECT?
Answer:
(I) only

Question 35.
In the figure find tan x.
Answer:
\(\frac{15}{8}\)

Question 36.
From the figure, find sin C.
Answer:

\(\frac{\mathrm{X}}{\mathrm{Z}}\)

Question 37.
Statement (A): sin² 67° + cos² 67° = 1
Statement (B): For any value of θ, sin₂ θ + cos₂ θ = 1
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true

Question 38.
Statement (A) : sin 47° = cos 43°.
Statement (B): sin θ = cos(90 + θ),
where θ is an acute angle.
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false, (ii)
Answer:
ii) A is true, B is false

Question 39.
In ΔABC, ∠B = 90°, AB = 3 cm and BC = 4 cm, then match the column.

A) sin C	i) 3/5
B) tan A	ii) 4/5
C) cos C	iii) 5/3
D) sec A	iv) 4/3

Answer:
A – (i), B – (iv), C – (ii), D – (iii)

A) sin C	i) 3/5
B) tan A	iv) 4/3
C) cos C	iii) 5/3
D) sec A	iv) 4/3

Question 40.
Match the following.

A) tan (90 – A)	i) sin 2A
B) \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)	ii) cos2A . sin2A
C) cos (90 – A)	iii) sin A
D) \(\frac{2 \tan \mathrm{A}}{1+\tan ^2 \mathrm{~A}}\)	iv) cot A

Answer:
A – (iv), B – (ii), C – (iii), D – (i)

A) tan (90 – A)	iv) cot A
B) \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)	ii) cos2A . sin2A
C) cos (90 – A)	iii) sin A
D) \(\frac{2 \tan \mathrm{A}}{1+\tan ^2 \mathrm{~A}}\)	i) sin 2A

Question 41.
Match the following.

A) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\)	i) cosec A + cot A
B) \(\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}\)	ii) 2sec A
C) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)	iii) sec A + tan A
D) \(\frac{\sin ^2 A}{1-\cos A}\)	iv) \(\frac{1+\sec A}{\sec A}\)

Answer:
A – (ii), B – (i), C – (iii), D – (iv)

A) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\)	ii) 2sec A
B) \(\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}\)	i) cosec A + cot A
C) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)	iii) sec A + tan A
D) \(\frac{\sin ^2 A}{1-\cos A}\)	iv) \(\frac{1+\sec A}{\sec A}\)

Question 42.
If sin θ = \(\frac{7}{25}\), then

A) cos θ	i) 24/25
B) cosec θ	ii) 7/24
C) tan θ	iii) 25/7
D) sec θ	iv) 25/24

Answer:
A – (i), B – (iii), C – (ii), D – (iv)

A) cos θ	i) 24/25
B) cosec θ	iii) 25/7
C) tan θ	ii) 7/24
D) sec θ	iv) 25/24

Question 43.
Match the following.

A) sin2 37° + sin2 53° + sin2 90°	i) 0
B) tan 35°.tan 45°.tan 55°	ii) 3
C) \(\frac{\sec 72^{\circ} \cdot \sin 18^{\circ}+\tan 72^{\circ} \cdot \cot 18^{\circ}}{\cos 60^{\circ}}\)	iii) 1
D) tan 60°/tan 30°	iv) 2
E) sin2 30° + cos2 30° – sin2 60°- cos2 60°	v) 4

Answer:
A – (iv), B – (iii), C – (v), D – (ii), E – (i)

A) sin2 37° + sin2 53° + sin2 90°	iv) 2
B) tan 35°.tan 45°.tan 55°	iii) 1
C) \(\frac{\sec 72^{\circ} \cdot \sin 18^{\circ}+\tan 72^{\circ} \cdot \cot 18^{\circ}}{\cos 60^{\circ}}\)	v) 4
D) tan 60°/tan 30°	ii) 3
E) sin2 30° + cos2 30° – sin2 60°- cos2 60°	i) 0

Question 44.
If x = 2019°, then what is the value of sin² x + cos² x ?
Solution:
If x = 2019°, then
sin²x + cos²x = sin²2019° + cos²2019°
= 1 [v sin²θ + cos²θ = 1]

Question 45.
If x is in first quadrant and sin x = cos x, then what is the value of x?
Sol.
Given, sin x = cos x
We know, sin (90°- θ) = cos θ
So, cos x = sin(90° – x)
⇒ sin x = sin(90° – x)
[note : If sin A = sin B, then A = B]
⇒ x = 90° – x
⇒ 2x = 90°
∴ x = 45°

Question 46.
tan θ =
A) \(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\)
B) \(\frac{1}{\sec \theta}\)
C) \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
D) \(\frac{1}{\sqrt{1+\cot ^2 \theta}}\)
Answer:
C) \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)

Question 47.
If sin θ = cos θ, then the value of sin 2θ = _________
Answer:
2 sin²θ

Question 48.
Assertion : tan 1° tan 2° tan 3° ……………… tan 89° = 1
Reasoning : In ΔABC if A + B = 90°, then tan A = cot B.
A) Both A and R are true and R is correct explanation of A.
B) Both A and R are true but R is not correct explanation of A
C) A is true but R is false
D) A is false but R is true
Answer:
A) Both A and R are true and R is correct explanation of A.

Question 49.
Which of the following is not true ?
(A) sin (90° – θ) = cosec θ
(B) sin² θ + cos² θ = 1
(C) cosec θ. sin θ = 1
(D) sin 90° = 1
Answer:
A) sin (90° – θ) = cosec θ

Question 50.
Match the following :

A) tan θ =	i) \(\frac{\cos \theta}{\sin \theta}\)
B) cot θ =	ii) \(\sqrt{1+\cot ^2 \theta}\)
C) cosec θ =	iii) \(\sqrt{\sec ^2 \theta-1}\)

a) A → (i), B → (ii), C → (iii)
b) A → (ii), B → (iii), C → (i)
c) A → (iii), B → (i), C → (ii)
d) A → (ii), B → (i), C (iii)
Answer:
c) A → (iii), B → (i), C → (ii)

A) tan θ =	iii) \(\sqrt{\sec ^2 \theta-1}\)
B) cot θ =	i) \(\frac{\cos \theta}{\sin \theta}\)
C) cosec θ =	ii) \(\sqrt{1+\cot ^2 \theta}\)

Question 51.
Write the relation between cos θ and sec θ.
Answer:
sec θ = \(\frac{1}{\cos \theta}\)

Question 52.
Value of cot² θ – cosec² θ is _________
Answer:
cot² θ – cosec² θ = -1

Question 53.
Match the following :

P) \(\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}\)	i) sec θ
q) tan θ cosec θ cot θ	ii) cosec θ
r) cosec (90° – θ)	iii) tan θ

A) p → i, q → iii, r → ii
B) p → iii, q → ii, r → i
C) p → ii, q → i, r → iii
D) p → iii, q → i, r → ii
Answer:
B) p → iii, q → ii, r → i

P) \(\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}\)	iii) tan θ
q) tan θ cosec θ cot θ	ii) cosec θ
r) cosec (90° – θ)	i) sec θ

Question 54.
Match the following :

P) sin θ	i) \(\frac{1}{\sec \theta}\)
Q) cos θ	ii) \(\sqrt{\sec ^2 \theta-1}\)
R) tan θ	iii) \(\sqrt{\frac{\sec ^2 \theta-1}{\sec ^2 \theta}}\)

Choose the correct answer.
A) P → (i), Q (ii), R → (iii)
B) P → (iii) Q → (i), R → (ii)
C) P → (iii), Q → (ii), R → (i)
D) P → (i), Q → (iii), R → (ii)
Answer:
B) P → (iii) Q → (i), R → (ii)

P) sin θ	iii) \(\sqrt{\frac{\sec ^2 \theta-1}{\sec ^2 \theta}}\)
Q) cos θ	i) \(\frac{1}{\sec \theta}\)
R) tan θ	ii) \(\sqrt{\sec ^2 \theta-1}\)

10th Class Maths Introduction to Trigonometry 2 Marks Important Questions

Question 1.
Find the value of x of 2 sin 3x = √3.
Solution:
Given 2 sin 3x = √3
sin 3x = \(\frac{\sqrt{3}}{2}\) = sin 60°
3x = 60° (∵ sin 60° = \(\frac{\sqrt{3}}{2}\))
x = \(\frac{60 }{3}\) = 20°

Question 2.
Find θ, if sin (θ + 36) = cos θ where θ + 36 is an acute angle.
Solution:
Given sin (θ + 36) = cos θ
sin (θ + 36) = sin (90 – θ)
θ + 36 = 90 – θ
θ + θ = 90 – 36 = 54
20 = 54
θ = \(\frac{\sqrt{54}}{2}\) = 27°
∴ θ = 27°

Question 3.
Prove that , \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = sec θ – tan θ.
Solution:
LHS = \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\)

= sec θ – tan θ = RHS

Question 4.
Prove that cot⁴A – 1 = cosec⁴A – 2 cosec²A.
LHS = cot⁴A – 1
= (cot²A)² – 1
We know cot²A = cosec²A – 1
cot²A – 1 = (cosec²A – 1)² – 1
= (cosec²A)² – 2cosec²A + 1 – 1
= cosec⁴A – 2cosec²A = RHS

Question 5.
Prove that sin⁴A – cos⁴A = 2sin²A – 1 = 1 – 2cos²A.
Solution:
LHS = sin⁴A – cos⁴A
= (sin²A)² – (cos²A)²
= (sin²A – cos²A) (sin²A + cos²A)
= sin²A – cos²A (∵ sin²A + cos²A = 1)
= sin²A – (1 – sin²A) (∵ cos²A = 1 – sin²)
= sin²A – 1 + sin²A
= 2sin²A – 1 = RHS
= 2(1 – cos²A) – 1 (∵ sin²A= 1 – cos²A)
= 2 – 2cos²A – 1
= 1 – 2cos²A = RHS

Question 6.
Prove that tan²θ.cos²θ = 1 – cos²θ.
Solution:
LHS = tan²θ.cos²θ (∵ tanθ = \(\frac{\sin \theta}{\cos \theta}\))
= \(\frac{\sin ^2 \theta}{\cos ^2 \theta}\).cos²θ (∵ sin²θ = 1 – cos²θ
= sin²θ = 1 – cos²θ) = RHS

Question 7.
If tan θ = \(\frac{3}{4}\), find the value of \(\frac{1-\cos \theta}{1+\cos \theta}\)
Solution:
Given tan θ = \(\frac{3}{4}\), then cos θ = \(\frac{4}{5}\)

Question 8.
If √3 tan θ = 3 sin θ, find the value of sin²θ – cos²θ.
Solution:
Given √3 tan θ = 3 sin θ
\(\frac{\tan \theta}{\sin \theta}\) = \(\frac{3}{\sqrt{3}}\)
\(\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta}{1}}\) = \(\frac{3}{\sqrt{3}}\) ×  \(\frac{\sqrt{3}}{\sqrt{3}}\)
\(\frac{\sin \theta}{\cos \theta}\) × \(\frac{1}{\sin \theta}\) = √3
cos = \(\frac{1}{\sqrt{3}}\)
sin² θ – cos² θ [∵ sin² θ = 1 – cos² θ]
= 1 – cos² θ – cos² θ
= 1 – 2cos² θ
= 1 – 2(\(\frac{1}{\sqrt{3}}\))²
= \(\frac{1}{1}\) – \(\frac{2}{3}\) = \(\frac{3-2}{3}\) = \(\frac{1}{3}\)
Therefore, sin²θ – cos²θ = \(\frac{1}{3}\).

Question 9.
Find the area of quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius = 4 cm
Circumference C = 2πr = 22 cm
= 2 × \(\frac{22}{7}\) × r = 22
r = \(\frac{22 \times 7}{2 \times 22}\) = \(\frac{7}{2}\) cm
Area of a quadrant = \(\frac{1}{4}\) πr²
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
Therefore, area of quadrant = \(\frac{77}{8}\) = 9.625 cm²

Question 10.
If a cosθ + bsinθ = m and asinθ – bcosθ = n then show that a² + b² = m² + n².
Solution:
acosθ + b sinθ = m
(a cosθ + b sinθ)² = m²
a² cos²θ + 2ab cosθ sinθ + b² sin²θ = m² …….. (1)
a sinθ – b cosθ = n
(asinθ – bcosθ)² = n²
a² sin²0 – 2ab sin0 cos0 + b² cos²0 = n² ……… (2)
(1) + (2) ⇒
a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = m² + n²
a² (1) + b² (1) = m² + n²
a² + b² = m² + n².

Question 11.
Prove that
\(\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}\) = 2cosec A.
Solution:

= 2cosec A = RHS
∴ LHS = RHS.

Question 12.
Evaluate: sin² 60° + 2tan 45° – cos² 30°
Solution:
sin² 60° + 2 tan 45° – cos² 30
= (\(\frac{\sqrt{3}}{2}\))² + 2(1)² – (\(\frac{\sqrt{3}}{2}\))² = 2.

Question 13.
Find the value of \(\frac{5}{\cot ^2 30}\) + \(\frac{1}{\sin ^2 30}\) – cot²45° + 2sin² 90°
Solution:
\(\frac{5}{\cot ^2 30}\) + \(\frac{1}{\sin ^2 30}\) – cot²45°+ 2sin² 90°
\(\frac{5}{(\sqrt{3})^2}+\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2}\) – 1² + 2(1)²
= \(\frac{5}{3}\) + \(\frac{4}{3}\) – 1 + 2 = \(\frac{9}{3}\) + 1 = 4.

Question 14.
If sinθ = cosθ then prove that tan²θ + cot²θ – 2 = 0.
Solution:
Given sin θ = cos θ
θ is acute, θ = 45°
tan²θ + cot²θ – 2 = tan² 45° + cot² 45° – 2
= 1² + 1² – 2 = 2 – 2 = 0.

Question 15.
Evaluate : \(\frac{\sin 30^{\circ}+\tan 45^{\circ}}{\sec 30^{\circ}+\cot 45^{\circ}}\)
Solution:

Question 16.
In a ΔADC, right angle at D, AC = 1.5 cm, DC = 3 cm; Find
i) tanθ,
ii) secθ + cosecθ.
Solution:

DC² = AC² + AD²
3² = (1.5)² + AD²
9 = 2.25 + AD²
AD² = 9 – 2.25 9
AD² = 9 – \(\frac{9}{4}\)
AD² = 9(1 – \(\frac{1}{4}\))
AD² = 9(1 – \(\frac{1}{4}\)

Question 17.
Find the value of x, if 2 sinx = √3.
Solution:
2 sinx = √3
sin x = \(\frac{\sqrt{3}}{2}\)
sin x = sin 60°
∴ x = 60°

Question 18.
Evaluate
(i) cos 76° – sin 14°, (ii) \(\frac{\tan 73^{\circ}}{\cot 17^{\circ}}\)
Solution:
To find the values of the following :
i) cos 76° – sin 14°
We can write cos 76 as cos (90 – 14)
∴ cos 76 = cos (90 – 14) = sin 14
(∵ cos (90 – θ) = sin θ)
∴ cos 76° – sin 14°
= sin 14° – sin 14° = 0 ________ (1)

ii) \(\frac{\tan 73^{\circ}}{\cot 17^{\circ}}\) = \(\frac{\tan (90-17)}{\cot 17}\)
= \(\frac{\cot 17^{\circ}}{\cot 17^{\circ}}\) = 1 (∵ tan (90 – θ) = cot θ)

Question 19.
Find the value of sin² 30° + cos² 60°.
Solution:
sin 30° = \(\frac{1}{2}\), cos 60° = \(\frac{1}{2}\)
sin² 30° + cos² 60° = (\(\frac{1}{2}\))² + (\(\frac{1}{2}\))²
= \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
∴ sin² 30° + cos² 60° = \(\frac{1}{2}\)

Question 20.
If sin (A + B) = 1 and cos B = \(\frac{1}{2}\), then find ∠A and ∠B. (0° < A + B ≤ 90°)
Solution:
sin (A + B) = 1
sin (A + B) = sin 90°
A + B = 90°
cos B = \(\frac{1}{2}\)
cos B = cos 60°
∴ ∠B = 60°
∴ ∠A = 90° – 60° = 30°

Question 21.
Simplify cot²θ – \(\frac{1}{\sin ^2 \theta}\)
Solution:
\(\frac{\cos ^2 \theta}{\sin ^2 \theta}\) – \(\frac{1}{\sin ^2 \theta}\)
= \(\frac{\cos ^2 \theta-1}{\sin ^2 \theta}\) = \(\frac{-\left(1-\cos ^2 \theta\right)}{\sin ^2 \theta}\) = \(\frac{-\sin ^2 \theta}{\sin ^2 \theta} \) = -1

Question 22.
Does sin θ = \(\frac{5}{3}\) exist for an acute angle θ? Give reason.
Solution:
Given ‘θ’ is acute ⇒ 0° < θ < 90°
So sin 0° = 0 and sin 90° = 1
So for 0° < θ < 90°,
sin θ value lies in between zero and one.
So sin θ value cannot be greater than 1.
So sinθ = \(\frac{5}{3}\) does not exist.

Question 23.
If sin x = \(\frac{3}{4}\), then what is the value of cosec x ?
Solution:

If sin x = \(\frac{3}{4}\)
cosec x = \(\frac{1}{\sin x}\) = \(\frac{1}{\frac{3}{4}}\) = \(\frac{4}{3}\)

Question 24.
Among sin 90°, cos 90°, tan 90°, cot 90°, sec 90° and cosec 90°; which is/are not defined ?
Solution:
sin 90° = 1
cos 90° = 0
tan 90° = not defined
cot 90° = 0
sec 90° = not defined
cosec 90° = 1
∴ tan 90°, sec 90° are not defined.

Question 25.
Express (sec² x – 1) (cot² x).
Solution:
(sec² x – 1) (cot² x)
= (tan² x) (cot² x) [∵ sec²A – tan²A = 1 sec²A – 1 = tan² A]
= \(\frac{\sin ^2 x}{\cos ^2 x} \cdot \frac{\cos ^2 x}{\sin ^2 x}\) [∵ tan A = \(\frac{\sin A}{\cos A}\) cot A = \(\frac{\cos A}{\sin A}\)
= 1

Question 26.
From the below figure, AB = 5 cm and AC = 13 cm, then calculate BC.

Solution:
In ΔABC, AB² + BC² = AC²
⇒ 5² + BC² = 13²
⇒ 25 + BC² – 169
⇒ BC² = 169 – 25
⇒ BC² = 144 ⇒ BC² = (12)²
∴ BC = 12 cm

Question 27.
Find the quadratic polynomial whose zeores are sin0° and tan45°.
Solution:
sin0° = 0, tan 45° = 1
Sum of zeroes = 0 + 1 = 1
Product zeroes = 0 x 1 = 0
Quadratic polynomial = x² – (sum of zeroes)x + product of zeroes
= x² – (1)x + 0 = x² – x

Question 28.
If tan θ = \(\frac{3}{4}\) then find sec θ.
Solution:

Question 29.
Show that tan² θ – \(\frac{1}{\cos ^2 \theta}[latex] = -1
Solution:

* Method – II : tan² θ – sec² θ
= – (sec² θ – tan² θ) = – 1

* Method – III : tan² θ – [latex]\frac{1}{\cos ^2 \theta}\)
= \(\frac{\sin ^2 \theta}{\cos ^2 \theta}\) – \(\frac{1}{\cos ^2 \theta}\)
= \(\frac{\sin ^2 \theta-1}{\cos ^2 \theta}\) – \(\frac{\sin ^2 \theta-1}{1-\sin ^2 \theta}\) = -1

Question 30.
If tan θ = √3 (θ is acute angle) then find the value of 1 + cos θ.
Solution:
tan θ = √3 = tan 60 (∵ θ is acute)
⇒ θ = 60
⇒ 1 + cos θ = 1 + cos 60 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
∴ 1 + cos θ = \(\frac{3}{2}\)

Question 31.
Evaluate: \(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}\) + \(\frac{\tan 42^{\circ}}{\cot 48^{\circ}}\).
Solution:
58°, 32° and 42°, 48° are complementary angles.
\(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}\) + \(\frac{\tan 42^{\circ}}{\cot 48^{\circ}}\)
= \(\frac{\sin 58^{\circ}}{\cos (90-58)^{\circ}}\) + \(\frac{\tan 42^{\circ}}{\cot (90-42)^{\circ}}\)
= \(\frac{\sin 58^{\circ}}{\sin 58^{\circ}}\) + \(\frac{\tan 42^{\circ}}{\tan 42^{\circ}}\) = 1 + 1 = 2

Question 32.
If sin A = \(\frac{1}{\sqrt{2}}\) and cot B = 1, prove that sin (A + B) = 1, where A and B both are acute angles.
Solution:
sin A = \(\frac{1}{\sqrt{2}}\) = sin 45° ⇒ A = 45°
cot B = 1 = cot 45° ⇒ B = 45°
L.H.S sin (A + B) = sin(45° + 45°)
= sin 90°
= 1 = R.H.S
∴ L.H.S = R.H.S

Question 33.
Express cos θ in terms of tan θ.
Solution:
cos θ = \(\frac{1}{\sec \theta}\) = \(\frac{1}{\sqrt{1+\tan ^2 \theta}}\).

Question 34.
If cos θ = \(\frac{1}{\sqrt{2}}\), then find the value of 4 + cot θ
Solution:
cos θ = \(\frac{1}{\sqrt{2}}\) cos 45 ⇒ θ = 45°
4 + cot θ = 4 + cot 45° = 4 + 1 = 5.

Question 35.
Is it correct to say that
sin θ = cos (90 – θ) ? why ?
Solution:
Yes.
∵ sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) (sin θ = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}\) )

Question 36.
Find the value of tan 2A, if cos 3A = sin 45°.
Solution:
cos 3A = sin 45°
cos 3A = \(\frac{1}{\sqrt{2}}\) = cos 45°
3A = 45° (∵ sin 45° = \(\frac{1}{\sqrt{2}}\))
A = \(\frac{45^{\circ}}{3}\) = 15°
tan 2A = tan (2 × 15°) = tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 37.
In a right triangle ABC, right angled at ‘C’ in which AB = 13 cm, BC = 5 cm, determine the value of cos² B + sin² A.
Solution:
We have, CosB = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{13}\),
Sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)

Cos²B + Sin² A
= \(\frac{25}{169}\) + \(\frac{25}{169}\) = \(\frac{25 + 25}{169}\) = \(\frac{50}{169}\).

Question 38.
Evaluate cosec 39° . sec 51° – tan 51°. cot 39°.
Solution:
sec A = cosec (90° – A)
sec 51° = cosec (90° – 51°) = cosec 39°
tan A = cot (90° – A)
tan 51° = cot (90° – 51°) = cot 39°
cosec 39° . sec 51° – tan 51° . cot 39°
= cosec 39°. cosec 39° – cot 39°. cot 39°
= cosec² 39° – cot² 39° (∵ cosec² A – cot² A = 1)
= 1.

Question 39.
Ravi says “the value of tan 0°.tan 1°. tan 2° ………. tan 89° is zero”. Do you agree with Ravi ? Give reason.
Solution:
tan 0°.tan(90 – 89)°.tan(90 – 88)° ……….. tan89°
tan 0°. cot 89° cot 88° …………… tan 88°.tan 89°
tan 0° (cot 89°. tan 89°) (cot 88. tan 88°) …….. tan 45°
= tan 0°. 1.1 1 = 0.1 = 0 (∵ tan0° = 0)
Yes. I agree with Ravi Answer.

Question 40.
Prove that
4 tan² 45° – cosec² 30° + cos² 30° = \(\frac{3}{4}\).
Solution:
4(1)² – (2)² + (\(\frac{\sqrt{3}}{2}\))²
= 4(1) – 4 + \(\frac{3}{4}\) = 4 – 4 + \(\frac{3}{4}\) = \(\frac{3}{4}\)

Question 41.
Using the figure given of AABC, prove that sin² 0 + cos² 0 = 1.

Solution:
From given figure,
sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) and cos θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
LHS = sin² θ + cos² θ
= (\(\frac{\mathrm{AB}}{\mathrm{AC}}\))² + (\(\frac{\mathrm{BC}}{\mathrm{AC}}\))²
= \(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\) + \(\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) = \(\frac{\mathrm{AB}^2+\mathrm{BC}^2}{\mathrm{AC}^2}\)
= \(\frac{\mathrm{AC}^2}{\mathrm{AC}^2}\) [∵ AB² + BC² = AC²]
= 1 = RHS
Hence it is proved.

Question 42.
Express tan θ in terms of sin θ.
Solution:
tan θ = \(\frac{\sin \theta}{\cos \theta}\)
sin² θ + cos² θ = 1
cos² θ = 1 – sin²θ
cos θ = \(\sqrt{1-\sin ^2 \theta}\)
∴ tan θ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)

Question 43.
Solve the quadratic equation 2 sin²θ – 3 sinθ + 1 = 0,
where 0° < θ ≤ 90°.
Solution:
2 sin²θ – 3 sinθ + 1 = 0
where 0° < θ ≤ 90°
2 sin²θ – 2sin θ – sinθ + 1 = 0
2 sinθ (sin θ – 1) – 1(sinθ – 1) = 0
(2 sin θ – 1)(sinθ – 1) = 0

2 sin – θ = 1
sin θ = \(\frac{1}{2}\)
sin θ = \(\frac{1}{2}\) = sin 30°
θ = 30°

sinθ – 1 = 0
sin θ = 1
sin θ = 1 = sin 90°
θ = 90°

Question 44.
Express ‘sin θ’ in terms of ‘tan θ’.
Solution:

10th Class Maths Introduction to Trigonometry 4 Marks Important Questions

Question 1.
Find acute angles A and B, if sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) and cos (A + 4B) = 0, A > B
Solution:
Given sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) and cos (A + 4B) = 0
We know,
sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) = sin 60°
A + 2B = 60° → (1)
cos (A + 4B) = 0 = cos 90°
A + 4B = 90° → (2)

Put B = 15° in (1)
⇒ A + 2 (15°) = 60°
∴ A = 60° – 30° = 30°
Therefore, A = 30° and B = 15°

Question 2.
If A and B are acute angles such that tan A = \(\frac{1}{2}\), tan B = \(\frac{1}{3}\). Find the value of \(\frac{\tan A+{tan} B}{1-\tan A \cdot \tan B}\).
Solution:
Given tan A = \(\frac{1}{2}\), tan B = \(\frac{1}{3}\)

\(\frac{5}{6}\) × \(\frac{6}{5}\) = 1
Therefore, \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1

Question 3.
If A = 30° and B = 60° verify that sin (A + B) = sin A cos B + cos A sin B and what do you observe ?
Solution:
Given A = 30°, B = 60°
LHS = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
RHS = sin A cos B + cos A sin B
= sin 30° cos 60° + cos 30° sin 60°
From the trigonometric ratio value table,
sin 30° = \(\frac{1}{2}\), cos 60° = \(\frac{1}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\), sin 60° = \(\frac{\sqrt{3}}{2}\)
= \(\frac{1}{2}\).\(\frac{1}{2}\) +\(\frac{\sqrt{3}}{2}\).\(\frac{\sqrt{3}}{2}\) = \(\frac{1}{4}\) + \(\frac{3}{4}\) = \(\frac{1 + 3}{4}\) = \(\frac{4}{4}\)
So, LHS = RHS
From this we get,
sin (A + B) = sin A cos B + cos A sin B

Question 4.
If θ is an acute angle and sin θ = cos θ, find the value of 2 tan²θ + sin²θ – 1.
Given sin θ = cos θ
Divide by cos θ as both sides,
\(\frac{\sin \theta}{\cos \theta}\) = \(\frac{\cos \theta}{\sin \theta}\)
tan θ = 1 = tan 45°
Therefore, θ = 45°
Then 2 tan²θ + sin²θ – 1
= 2 (tan 45°)² + (sin 45°)² – 1
= 2.1² + (\(\frac{1}{\sqrt{3}}\))² – 1
= 2 + \(\frac{1}{2}\) – 1
= \(\frac{4 + 1 – 2}{2}\) = \(\frac{5 – 2}{2}\) = \(\frac{3}{2}\)
∴ 2 tan²θ + sin²θ – 1 = \(\frac{3}{2}\)

Question 5.
Prove that cos²θ + \(\frac{1}{1+\cot ^2 \theta}\) = 1.
Solution:
LHS = cos²θ + \(\frac{1}{1+\cot ^2 \theta}\)
we have 1 + cot²θ = cosec²θ
= cos²θ + \(\frac{1}{{cosec}^2 \theta}\)
= cos²θ + sin²θ (∵ \(\frac{1}{{cosec} \theta}\) = sinθ)
= 1 = RHS (∵ cos²θ + sin²θ = 1)
Hence cos²θ + \(\frac{1}{1+\cot ^2 \theta}\) = 1

Question 6.
Prove that \(\frac{1}{1+\sin \theta}\) + \(\frac{1}{1-\sin \theta}\) = 2 sec²θ
Solution:
LHS = \(\frac{1}{1+\sin \theta}\) + \(\frac{1}{1-\sin \theta}\)
= \(\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}\)
2.\(\frac{1}{1-\sin ^2 \theta}\) (∵ 1 – sin²θ = cos²θ and \(\frac{1}{\cos \theta}\) = secθ)
= 2.\(\frac{1}{\cos ^2 \theta}\) = 2 sec²θ = RHS
Hence \(\frac{1}{1+\sin \theta}\) + \(\frac{1}{1-\sin \theta}\) = 2 sec²θ

Question 7.
(1 + tan²θ) (1 + sin θ) (1 – sin θ) = 1.
Solution:
LHS = (1 + tan²θ) (1 + sin θ) (1 – sin θ)
= (1 + tan²θ) (1 – sin²θ)
We have,
1 + tan²θ = sec²θ, 1 – sin²θ = cos²θ
= sec²θ.cos²θ (∵ sec θ – cos θ = 1)
= (sec θ.cos θ)² = 1² = 1 = RHS.
Hence, (1 + tan²θ) (1 + sin θ) (1 – sin θ) = 1.

Question 8.
Show that \(\frac{\sin \theta}{1-\cos \theta}\) = cosec θ + cot θ.
Solution:

Question 9.
Prove that \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = cosec θ + cot θ.
Solution:

Question 10.
Prove that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Solution:
LHS = (cosec θ – cot θ)²
= (\(\frac{1}{\sin \theta}\) – \(\frac{\cos \theta}{\sin \theta}\)) (∵ cosec θ = \(\frac{1}{\sin \theta}\), cot = \(\frac{\cos \theta}{\sin \theta}\))

Question 11.
Prove that sec⁴θ – sec²θ = tan⁴θ + tan²θ.
Solution:
LHS = sec⁴θ – sec2θ = sec²θ (sec²θ – 1)
We know, sec²θ = 1 + tan²θ,
= (1 + tan²θ)(1 + tan²θ – 1)
= (1 + tan²θ) tan²θ
= tan²θ + tan⁴θ
= tan⁴θ + tan²θ = RHS
Hence, sec⁴θ – sec²θ = tan⁴θ + tan²θ

Question 12.
Prove that \(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\) = tan θ + cot θ.
Solution:
LHS = \(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\)
We have, sec²θ = 1 + tan²θ, cosec²θ = 1 + cot²θ
= \(\sqrt{1+\tan ^2 \theta+1+\cot ^2 \theta}\)
= \(\sqrt{2+\tan ^2 \theta+\cot ^2 \theta}\)
= \(\sqrt{2 \cdot \tan \theta \cdot \cot \theta+\tan ^2 \theta+\cot ^2 \theta}\)
= \(\sqrt{(\tan \theta+\cot \theta)^2}\)
= tan θ + cot θ = RHS
Hence, \(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\) = tan θ + cot θ

Question 13.
Prove that sin⁴A + cos⁴A = 1 – 2 sin²A – cos²A.
Solution:
LHS = sin⁴A + cos⁴A
= (sin²A)2 + (cos²A)² + 2 sin²A cos²A – 2 sin²A cos²A
= (sin²A + cos²A)² – 2 sin²A cos²A
(∵ sin²A + cos²A = 1)
= 1 – 2 sin²A cos²A = RHS

Question 14.
Prove that sin⁶A + cos⁶A = 1 – 3 sin²A cos²A.
Solution:
LHS = sin⁶A + cos⁶A
= (sin²A)³ + (cos²A)³ [∵ a³ + b³ = (a + b) (a² + b² – 3ab)]
= (sin²A + cos²A) [(sin²A)² + (cos²A)² – sin²A.cos²A]
We have sin²A + cos²A = 1
= 1 [(sin²A)² + (cos²A)² + 2
sin²A.cos²A – 3sin²A.cos²A]
= (sin²A + cos²A)² – 3.sin²A.cos²A
= (1)² – 3 sin²A.cos²A
= 1 – 3 sin²A cos²A = RHS

Question 15.
Prove that \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}\) sin θ + cos θ = 1.
Solution:

Question 16.
If sin θ + sin²θ = 1, then prove that cos²θ + cos⁴θ = 1.
Solution:
Given sin θ + sin²θ = 1
sin θ = 1 – sin²θ = cos⁴θ
LHS = cos²θ + cos²θ
= cos²θ + (cos²θ)²
= cos²θ + (sin θ)²
= cos²θ + sin²θ = 1 = RHS
Hence, cos²θ + cos⁴θ = 1.

Question 17.
Prove that \(\frac{1+\sec \theta}{\sec \theta}\) = \(\frac{\sin ^2 \theta}{1-\cos \theta}\)
Solution:

Question 18.
Prove that \(\frac{\cot A-\cos A}{\cot A+\cos A}\) = \(\frac{{cosec} A-1}{{cosec} A+1}\).
Solution:

Question 19.
If θ is an acute angle and sin θ = cos θ, find the value of 2 tan²θ + sin²θ – 1.
Solution:
Given sin θ = cos θ
\(\frac{\sin \theta}{\cos \theta}\) = 1
tan θ = 1 = tan 45°
∴ θ = 45°
Then 2 tan²θ + sin²θ – 1
= 2 tan²45° + sin²45° – 1
= 2(1)² + (\(\frac{1}{\sqrt{2}}\))² – 1 = 2 + \(\frac{1}{2}\) – 1
1 \(\frac{1}{2}\) (or) \(\frac{3}{2}\)
∴ 2 tan²θ + sin²θ – 1 = \(\frac{3}{2}\)

Question 20.
If θ = 30° then verify tan 2θ = \(\frac{2 \tan \theta}{1-\tan ^2 \theta}\).
Solution:
Given θ = 30°
LHS = tan 2θ = tan (2.30°) = tan 60° = √3

Question 21.
Prove that \(\frac{\sin A+\cos A}{\sin A-\cos A}\) = \(\frac{\sin A-\cos A}{\sin A+\cos A}\) = \(\frac{2}{\sin ^2 A-\cos ^2 A}\) = \(\frac{2}{2 \sin ^2 A-1}\)
Solution:
\(\frac{\sin A+\cos A}{\sin A-\cos A}\) = \(\frac{\sin A-\cos A}{\sin A+\cos A}\)
= \(\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)}\)

Question 22.
Prove that tan²θ + cot²θ + 2 = sec²θ.cosec²θ.
Solution:
LHS = tan²θ + cot²θ + 2
= tan²θ + cot²θ + 2.tan θ.cot θ
= (tan θ + cot θ)²
= (\(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\))²
= (\(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cdot \cos \theta}\))²
= (\(\frac{1}{\sin \theta \cdot \cos \theta}\))² = \(\frac{1}{\sin ^2 \theta}\).\(\frac{1}{\cos ^2 \theta}\)
= cosec²θ.sec²θ = RHS
LHS = RHS
Hence, tan²θ + cot²θ + 2 = cosec²θ.sec²θ

Question 23.
Prove that (1 + cot A – cosec A) (1 + tan A + sec A) = 2.
Solution:
LHS = (1 + cot A – cosec A) (1 + tan A + sec A)

Hence, (1 + cot A – cosec A) (1 + tan A + sec A) = 2

Question 24.
If tan (A + B) = 1 and cos (A – B) = \(\frac{\sqrt{3}}{2}\), 0° < A + B < 90, A > B then find values of A and B.
Solution:
tan (A + B) = 1 = tan 45°
∴ A + B = 45° ______ (1)
cos (A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ A – B = 30° _______ (2)
Solving the equation (1) and (2) we get

then A + B = 45
37.5 + B = 45
⇒ B = 45 – 37.5 = 7.5
So A = 37.5°, B = 7.5‘

Question 25.
If x = a sec θ and y = b tan θ, then prove that \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1.
Solution:
\(\frac{x}{a}\) = sec θ, \(\frac{y}{b}\) = tan θ
sec² θ = tan² θ = 1
(\(\frac{x}{a}\))² – (\(\frac{y}{b}\))² = 1 ⇒ \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1

Question 26.
If A, B, C are interior angles of ΔABC, then show that
sin (\(\frac{A+B}{2}\)) + cos (\(\frac{A+B}{2}\)) = cos \(\frac{C}{2}\) + sin \(\frac{C}{2}\)
Solution:
A + B + C = 180°
A + B = 180° – C
(\(\frac{A+B}{2}\)) = 90° – \(\frac{C}{2}\)
sin (\(\frac{A+B}{2}\)) = sin(90° – \(\frac{C}{2}\)) = cos \(\frac{C}{2}\) …….. (1)
cos (\(\frac{A+B}{2}\)) = cos(90° – \(\frac{C}{2}\)) = sin \(\frac{C}{2}\) …….. (2)
(1) + (2)
= sin (\(\frac{A+B}{2}\)) + cos (\(\frac{A+B}{2}\)) = cos \(\frac{C}{2}\) + sin \(\frac{C}{2}\)

Question 27.
Prove that \(\sqrt{\frac{{cosec} A+1}{{cosec} A-1}}\) – \(\sqrt{\frac{{cosec} A-1}{{cosec} A+1}}\) = 2 cot A.
Solution:
\(\sqrt{\frac{{cosec} A+1}{{cosec} A-1}}\) – \(\sqrt{\frac{{cosec} A-1}{{cosec} A+1}}\)
= \(\frac{{cosec} A+1-({cosec} A-1)}{\sqrt{{cosec}^2 A-1}}\)
= \(\frac{{cosec} A+1-{cosec} A+1}{\cot A}\)
= \(\frac{2}{\cot A}\) = 2 tan A.

Question 28.
Find measure of the angles A and B, if cos (A – B) = \(\frac{\sqrt{3}}{2}\) and sin (A + B) = \(\frac{\sqrt{3}}{2}\).
Solution:
cos (A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30°
A – B = 30° ……………. (1)
sin (A + B) = \(\frac{\sqrt{3}}{2}\) = sin 60
A + B = 60° ………… (2)
(1) + (2) :
A + B = 60
A – B = 30
2A = 90°
∴ A = 45°; B = 15°.

Question 29.
Find the value \(\frac{\tan ^2 60^{\circ}+\cot ^2 30^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 60^{\circ}}\)
Solution:

Question 30.
Prove that : \(\frac{1}{\sin ^2 \theta}\) – cot²θ = 1
Solution:
Given : \(\frac{1}{\sin ^2 \theta}\) – cot²θ
= \(\frac{1}{\sin ^2 \theta}\) – \(\frac{\cos ^2 \theta}{\sin ^2 \theta}\) [∵ cot²θ = [∵ cot² \(\frac{\cos ^2 \theta}{\sin ^2 \theta}\)
= \(\frac{1-\cos ^2 \theta}{\sin ^2 \theta}\)
= \(\frac{\sin ^2 \theta}{\sin ^2 \theta}\) = 1 [∵ sin²θ + cos²θ = 1 ⇒ sin²θ = 1 – cos²θ
∴ \(\frac{1}{\sin ^2 \theta}\) – cot²θ

Question 31.
If cosec(A + B) = 1 and cot (A – B) = √3 , 0° < A + B < 90°, A > B, then find A and B.
Solution:
cosec (A + B) = 1 = cosec 90°
A + B = 90° ………….. (1)
cot (A – B) = √3 = cot 30°
A – B = 30° ……………. (2)
(1) + (2) ⇒
A + B = 90°
A – B = 30°
2A = 120°
⇒ A = \(\frac{120^{\circ}}{2}\) = 60°
A + B = 90°
60°+ B = 90°
B = 90° – 30° = 30°
∴ A = 60°, B = 30°.

Question 32.
Prove that
\(\frac{\sin \theta}{1+\cos \theta}\) + \(\frac{1+\cos \theta}{\sin \theta}\) = 2 cosec θ
Solution:

= \(\frac{2}{{Sin} \theta}\) = 2 Cosec θ = RHS
LHS = RHS. Hence proved

Question 33.
cos A = \(\frac{12}{13}\), then, find sin A and 13 tan A.
Solution:
Given, cos A = \(\frac{12}{13}\)
AC² = AB² + BC²
13² = 12² + x²
169 = 144 + x²
x² = 25 ⇒ x = 5
sin A = \(\frac{5}{13}\)

Question 34.
If A, B and C are interior angles of Δ ABC, then show that tan(\(\frac{B+C}{2}\)) = cos\(\frac{A}{2}\).
Solution:
Given A, B and C are interior angles of right angled triangle ABC.
Then A + B + C = 180°
On dividing the above equation by 2 on both sides, we get
\(\frac{A}{2}\) + \(\frac{B+C}{2}\) = 90°
∴ \(\frac{B+C}{2}\) = 90° – \(\frac{A}{2}\)
On taking tan ratio on both sides.
tan(\(\frac{B+C}{2}\)) = tan(90° – \(\frac{A}{2}\).
∴ tan(\(\frac{B+C}{2}\)) = cot \(\frac{A}{2}\) [v tan (90 – θ) = cot θ]
Hence proved.

Question 35.
The sides of a right angle triangle PQR are PQ = 7cm, QR = 25 cm and ∠P = 90° respectively, then find sin Q + sin R.
Solution:
Given that ΔPQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.

By Pythagoras theorem and
QR² = PQ² + PR²
∴ PR² = QR² – PQ²
= (25)² – (7)² = 625 – 49 = 576
∴ PR = \(\sqrt{576}\) = 24 cm
Sin Q = \(\frac{\mathrm{PR}}{\mathrm{RQ}}\) = \(\frac{1}{2}\); sin R = \(\frac{\mathrm{PQ}}{\mathrm{RQ}}\) = \(\frac{7}{25}\)
∴ sinQ + sinR = \(\frac{24}{25}\) + \(\frac{7}{25}\)
= \(\frac{24 + 7}{25}\) = \(\frac{31}{25}\)

Question 36.
If tan θ = \(\frac{5}{12}\), then find sec θ and cosec θ.
Solution:

Given tan θ = \(\frac{5}{12}\)
tan θ = \frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{5}{12}
Opposite side to angle θ is 5.
∴ BC = 5
Adjacent side to angle θ is 12.
∴ AC = 12
By Pythagoras theorem, in ΔABC
AB² = AC² + BC²
= (12)² + (5)² = 144 + 25 = 169
AB = \(\sqrt{169}\) = 13 = Hypotenuse
cosec θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) = \(\frac{13}{12}\)
sec \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{13}{12}\)

Question 37.
Express the following in terms of trigonometric ratios of angles between 0° and 45°.
i) sin 81° + tan 75°
ii) cos 65° + cot 75°
Solution:
i) sin 81° + tan 75°
= sin (90 – 9)° + tan (90 – 15)°
= cos 9° + cot 15°
{∵ sin (90 – θ) = cos θ ; tan (90 – θ) = cot θ]

ii) cos 65° + cot 75°
= cos (90 – 25)° + cot (90 – 15)°
= sin 25° + tan 15°
{∵ cos (90 – θ) = sin θ ; cot (90 – θ) = tan θ]

Question 38.
Show that \(\frac{\tan \theta}{1-\cot \theta}\) + \(\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
Solution:


= 1 + sec θ – cosec θ = RHS
∴ LHS = RHS.

Question 39.
Prove that \(\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}\) = tan θ.
Solution:
LHS = \(\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}\)

= tan A = RHS ∴ LHS = RHS.

10th Class Maths Introduction to Trigonometry 8 Marks Important Questions

Question 1.
If sin θ = \(\frac{12}{13}\), find the value of \(\frac{\sin ^2 \theta-\cos ^2 \theta}{2 \sin \theta \cdot \cos \theta}\) \(\frac{1}{\tan ^2 \theta}\).
Solution:

Question 2.
Evaluate \(\frac{2}{3}\) (cos⁴30° – sin⁴45°) –
3(sin²60° – sec²45°) + \(\frac{1}{4}\) cot30°.
Solution:
From the trigonometric ratio value table,
cos 30° = \(\frac{\sqrt{3}}{2}\), sin 45° = \(\frac{1}{\sqrt{2}}\)
sin 60° = \(\frac{\sqrt{3}}{2}\) , sec 45° = √2
cot 30° = √3
= \(\frac{2}{3}\) (cos⁴30° – sin⁴45°) – 3 (sin²60° – sec²45°) + \(\frac{1}{4}\) cot²30°

Question 3.
Prove that (sin θ + sec θ)² + (cos θ + cosec θ)² = (1 + sec θ.cose θ)².
Solution:
LHS = (sin θ + sec θ)² + (cos θ + cosec θ)²
= sin²θ + sec²θ + 2 sin θ.sec θ + cos²θ + cosec²θ + 2 cos θ.cosec θ

= 1 + \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \cdot \sin ^2 \theta}\) + 2(\(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \cdot \sin \theta}\))
= 1 + \(\frac{1}{\cos ^2 \theta \cdot \sin ^2 \theta}\) + 2.\(\frac{1}{\cos \theta \cdot \sin \theta}\)
= 1 + sec²θ.cosec²θ + 2.sec θ.cosecθ
= (1 + sec θ.cosec θ)² = RHS
Hence, (sin θ + sec θ)² + (cos θ + cosec θ)²
= (1 + sec θ.cosec θ)²

Question 4.
Prove that 2 sec²θ – sec⁴θ – 2 cosec²θ + cosec⁴θ = cot⁴θ – tan⁴θ.
Solution:
LHS = 2 sec²θ – sec²θ – 2 cosec²θ + cosec⁴θ
We have, sec²θ – 1 = tan²θ,
cosec²θ – 1 = cot²θ
= (cosec⁴θ – 2 cosec²θ + 1) – (sec⁴θ – 2 sec²θ + 1)
= (cosec²θ – 1)² – (sec²θ – 1)²
= (cot²θ)² – (tan²θ)²
= cot⁴θ – tan⁴θ = RHS
Hence, 2 sec²θ – sec⁴θ – 2 cosec²θ + cosec⁴θ = cot⁴θ – tan⁴θ.

Question 5.
Prove that \(\frac{\sin \theta}{\cot \theta+{cosec} \theta}\) = 2 + \(\frac{\sin \theta}{\cot \theta-{cosec} \theta}\).
Solution:

Question 6.
Prove that (cosec θ – sin θ) (sec θ – cos θ) = \(\frac{1}{\tan \theta+\cot \theta}\).
Solution:
LHS = (cosec θ – sin θ) (sec θ – cos θ)

Question 7.
\(\frac{\tan A}{1-\cot A}\) + \(\frac{\cot A}{1-\tan A}\) = 1 + tan A + cot A = 1 + sec A cosec A.
Solution:


= 1 + tan A + cot A = 1 + sec A cosec A.

Question 8.
Prove that 2(sin⁶θ + cos⁶θ) – 3 (sin⁴θ + cos⁴θ) + 1 = 0
Solution:
LHS = 2(sin⁶θ + cos⁶θ) – 3 (sin⁴θ + cos⁴θ) + 1 = 0
= 2[sin²θ)³ + cos²θ)³] – 3 [sin²θ)² + (cos²θ)²] + 1
We know, a³ + b³ = (a + b) (a² + b² – ab) and sin²θ + cos²θ = 1
= 2 (sin²θ + cos²θ) [(sin²θ)² + (cos²θ)² – sin²θ cos²θ] – 3 [(sin²θ)² + cos²θ)² + 2 sin²θ
cos²θ – sin²θcos²θ] + 1

Question 9.
Prove that (sin⁸θ – cos⁸θ) = (sin²θ – cos²θ) (1 – 2 sin²θ cos²θ).
Solution:

Question 10.
Prove that \(\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}\) + \(\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}\) = 2 sec θ.
Solution:

Question 11.
If sec θ = x + \(\frac{1}{4 x}\), then prove that sec θ + tan θ = 2x (or) \(\frac{1}{2 x}\).
Solution:
Given sec 0 = x + \(\frac{1}{4 x}\)
We know that, sec²θ – tan²θ = 1
tan²θ = sec²θ – 1

Question 12.
Evaluate 4 (sin⁴30° + cos⁴60°) – \(\frac{2}{3}\) (sin²60° – cos²45°) + \(\frac{1}{2}\)tan²60°.
Solution:
Given 4 (sin⁴30° + cos⁴60°) – \(\frac{2}{3}\) (sin²60° – cos²45°) + \(\frac{1}{2}\)
tan²60°.
We know, sin 30° = \(\frac{1}{2}\), cos 60° = \(\frac{1}{2}\),
cos 45° = \(\frac{1}{\sqrt{2}}\) sin60° = \(\frac{\sqrt{3}}{2}\)
tan 60° = √3

Question 13.
Prove that
tan²A – tan²B = \(\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 B \cdot \cos ^2 A}\) = \(\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}\).
Solution:
LHS = tan²A – tan²B
= \(\frac{\sin ^2 \mathrm{~A}}{\cos ^2 \mathrm{~A}}-\frac{\sin ^2 \mathrm{~B}}{\cos ^2 \mathrm{~B}}\)

Question 14.
Prove that (1 + tan A.tan B)² + (tan A – tan B)² = sec²A.sec²B.
Solution:
LHS = (1 + tan A.tan B)² + (tan A – tan B)²
= 1 + 2tanA tanB + tan²A.tan²B + tan²A + tan²B – 2tanA tanB
= 1 + tan²A.tan²B + tan²A + tan²B
= (1 + tan²A) + (tan²B + tan²A.tan²B)
= (1 + tan²A) + tan²B (1 + tan²A)
= (1 + tan²A) (1 + tan²B) (1 + tan²θ = sec²θ)
= sec²A.sec²B = RHS
Hence, (1 + tan A.tan B)² + (tan A – tan B)² = sec²A.sec²B

Question 15.
Prove that cot²A (\(\frac{\sec A-1}{1+\sin A}\)) + sec²A (\(\frac{\sin A-1}{1+\sec A}\)) = 0.
Solution:

Question 16.
Prove that \(\frac{\cos A}{1-\sin A}\) + \(\frac{\sin A}{1-\cos A}\) + 1 = \(\frac{\sin A \cdot \cos A}{(1-\sin A)(1-\cos A)}\).
Solution:

Question 17.
If tan θ + sin θ = m and tan θ – sin θ = n
show that m² – n² =4.\(\sqrt{\mathrm{mn}}\).
Given tan θ + sin θ = m
and tan θ – sin θ = n
[:(a + b)² – (a – b)2 = 4ab]
LHS m² – n²
=(tan θ + sin θ)² – (tan θ – sin θ)²
= 4 tan θ. sin θ
Now, RHS =4.\(\sqrt{\mathrm{mn}}\)

Question 18.
If \(\frac{\cos \alpha}{\cos \beta}\) = m and \(\frac{\cos \alpha}{\sin \beta}\) = n show that (m² + n²)cos²β = n²
Solution:
Given that \(\frac{\cos \alpha}{\cos \beta}\) = m and \(\frac{\cos \alpha}{\sin \beta}\) = n
LHS = (m² + n²)cos²β = n²

Question 19.
\(\frac{\sec \theta-1}{\sec \theta+1}\) + (\(\frac{\sin \theta}{1+\cos \theta}\))²
Solution:

Question 20.
If cot θ = \(\frac{9}{12}\), then find the value of \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) + \(\frac{\sec \theta+{cosec} \theta}{\sec \theta-{cosec} \theta}\).
Solution:
Given that cot θ = \(\frac{9}{12}\) => tan θ = \(\frac{12}{9}\)
Given that


Hence the given expression value is 14.

Question 21.
Prove that
(1 + tan²θ) + (1 + \(\frac{1}{\tan ^2 \theta}\)) = \(\frac{1}{\sin ^2 \theta-\sin ^4 \theta}\).
Solution:

(∵ 1 – sin²θ = cos²θ)
∴ L.H.S = R.H.S

Question 22.
If \(\frac{\cos \alpha}{\cos \beta}\) = m, \(\frac{\cos \alpha}{\sin \beta}\) = n, then show that (m² + n²) cos²α = m²n².
Solution:

Question 23.
If sec θ + tan θ = P, then prove that sin θ = \(\frac{\mathbf{P}^2-1}{\mathbf{P}^2+1}\).
Solution:
If sec θ + tan θ = P
Then sec θ – tan θ = \(\frac{1}{P}\) [∵ sec² θ – tan² θ = 1]

Question 24.
In an acute angled triangle ABC, if sin(A + B – C) = \(\frac{1}{2}\) and cos(B + C – A) =
\(\frac{1}{2}\), then find ∠A, ∠B and ∠C.
Solution:
sin(A + B – C) = \(\frac{1}{2}\) = sin 30°
⇒ A + B – C = 30° ……. (1)
cos(B + C – A) = \(\frac{1}{2}\) = cos 60°
⇒ B + C – A = 60° ……… (2)
On adding equations (1) and (2)

⇒ B = \(\frac{90}{2}\) = 45°
Substituting the value of B in equation (2)
C – A = 60° – 45° = 15° ………….. (3)
B + C + A = 180°
(sum of angles of a triangle)
C + A= 180°-45° = 135° ………….. (4)
Now solving equations (3) and (4)

∠A = 135° – 75° = 60°
∴ ∠A = 60°, ∠B = 45°, ∠C = 75°

AP 10th Class Maths Chapter 8 Important Questions Introduction to Trigonometry

Question 1.
If sin A = cos A, then find the value of A.
Solution:
If sin A = cos A
\(\frac{\sin A}{\cos A}=\frac{\cos A}{\cos A}\) = 1
⇒ tan A = 1 = tan 45° ⇒ A = 45°
(OR)
sin 45° = cos 45°
\(\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\)
So, A = 45°

Question 2.
Find the value of x, if 2 sinx = √3
Solution:
2 sinx = √3
sin x = \(\frac{\sqrt{3}}{2}\)
sin x = sin 60°
∴ x = 60°

Question 3.
Evaluate
(i) cos 76° – sin 14°,
(ii) \(\frac{\tan 73^{\circ}}{\cot 17^{\circ}}\)
Solution:
To find the values of the following :
i) cos 76° – sin 14°
We can write cos 76 as cos (90 – 14)
∴ cos 76 = cos (90 – 14) = sin 14
(∵ cos (90 – θ) = sin θ)
∴ cos 76° – sin 14°
= sin 14° – sin 14° = 0 ………… (1)

(ii) \(\frac{\tan 73^{\circ}}{\cot 17^{\circ}}\) = \(\frac{\tan (90-17)}{\cot 17}\)
= \(\frac{\cot 17^{\circ}}{\cot 17^{\circ}}\) = 1
(∵ tan (90 – θ) = cot θ

Question 4.
Find the value of tan²45° + cot² 30°.
Solution:
tan² 45° + cot²30°
= (1)² + (√3)²
= 1 + 3 = 4

Question 5.
Find the value of sin² 30° + cos² 60°.
sin 30° = \(\frac{1}{2}\),cos 60° = \(\frac{1}{2}\)
sin² 30° + cos² 60° = (\(\frac{1}{2}\))² + (\(\frac{1}{2}\))²
= \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
∴ sin² 30° + cos² 60° = \(\frac{1}{2}\)

Question 6.
If sin (A + B) = 1 and cos B = \(\frac{1}{2}\), then find ∠A and ∠B. (0° < A + B ≤ 90°)
Solution:
sin (A + B) = 1
sin (A + B) = sin 90°
A + B = 90°
cos B = \(\frac{1}{2}\)
cos B – cos 60°
∴ ∠B = 60°
∴ ∠A = 90° – 60° = 30°

Question 7.
Simplify cot²θ – \(\frac{1}{\sin ^{2} \theta}\)
Solution:

Question 8.
Does sinθ \(\frac{5}{3}\) exist for an acute angle θ?
Give reason.
Solution:
Given ‘θ’ is acute => 0° < θ < 90°
So sin 0° = 0 and sin 90° = 1
So for 0° < θ < 90°,
sin θ value lies in between zero and one.
So sin θ value cannot be greater than 1.
So sinθ = \(\frac{5}{3}\) does not exist.

Question 9.
If sin x = \(\frac{3}{4}\), then what is the value of cosec x?
Solution:
If sin x = \(\frac{3}{4}\)

Question 10.
Among sin 90°, cos 90°, tan 90°, cot 90°, sec 90° and cosec 90°; which is/are not defined ?
Solution:
sin 90° =1
cos 90° = 0
tan 90° = not defined
cot 90° = 0
sec 90° = not defined
cosec 90° = 1
∴ tan 90°, sec 90° are not defined.

Question 11.
Express (sec² x – 1) (cot² x).
Solution:
(sec² x – 1) (cot² x)
= (tan² x) (cot² x) [∵ sec²A – tan²A = 1 sec² A – 1 = tan² A]
= \(\frac{\sin ^{2} x}{\cos ^{2} x} \cdot \frac{\cos ^{2} x}{\sin ^{2} x}\) [∵ tan A = \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) cot A = \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\)]
= 1

Question 12.
Find ∠B, if tan (A – B) = \(\frac{1}{\sqrt{3}}\) and sinA = \(\frac{\sqrt{3}}{2}\). Also find cos B. (A, B < 90°)
Solution:
tan (A – B) = \(\frac{1}{\sqrt{3}}\)
∴ tan (A – B) = tan 30°
∴ A – B = 30°
sin A = \(\frac{\sqrt{3}}{2}\)
sin A = sin 60°
∴ A = 60°
substituting A – B = 30°
60° – B = 30°
B = 30°

Question 13.
If tan A = \(\frac{1}{\sqrt{3}}\) and tan B = √3 , then find sin A. cos B + cos A . sin B.
(A, B < 90°).
Solution:
Given, tan A = \(\frac{1}{\sqrt{3}}\) = and tan B = √3
Given A, B < 90°
We know \(\frac{1}{\sqrt{3}}\) = tan 30°
so tan A = \(\frac{1}{\sqrt{3}}\) = tan 30°
⇒ A = 30° …………….. (1)
and tan B = √3 = tan 60°
⇒ B = 30° …………….. (2)
then
sin A cos B + cos A sin B
= sin 30° cos 60° + cos 30° sin 60°
Which can be written as
sin (A + B) = sin (30+60) = sin 90 = 1
∴ sin A cos B + cos A sin B = 1
(or)
sin 30 . cos 60 + cos 30 . sin 60
= \(\frac{1}{2} \cdot \frac{1}{2}+\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\) = \(\frac{1}{4}+\frac{3}{4}=\frac{4}{4}\) = 1

Question 14.
Prove that
tan²A – sin²A = tan²A . sin²A.
Solution:
tan²A – sin²A
= \(\frac{\sin ^{2} A}{\cos ^{2} A}\) – sin²A
sin²A (\(\frac{1}{\cos ^{2} A}\) – 1 )
= sin²A (sec²A – 1)
= sin²A. tan²A

Question 15.
If cos A = \(\frac{7}{25}\), then find sin A and cosec A. What do you observe?
Solution:
Given that cos A = 7/25
sin² A = 1 – cos² A

I observed that sin A . cosec A = 1.

Question 16.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.
Solution:
tan 2A = cot(A – 18°)
= cot[90 – [90 – (A – 18°)]]
tan 2A = tan [90 – (A – 18°)]
2A = 90-(A – 18°) = 90 – A + 18°
⇒ 3A = 108°
∴ A = 36°

Question 17.
If 4 tan θ = 3, then find the value of sec θ and cosec θ.
Solution:
4 tan θ = 3 ⇒ tan θ = \(\frac{3}{4}\)

AC = \(\sqrt{\mathrm{BC}^{2}+\mathrm{AB}^{2}}\) = \(\sqrt{4^{2}+3^{2}}\) = 5
sec θ = \(\frac{5}{4}\); cosec θ = \(\frac{5}{3}\)

Question 18.
If tan 2A = cot (A – 27), where 2A is an acute angle, find the value of A.
Solution:
Given that 2A is an acute angle.
We know if tan = α = cot β (α, β acute) then α + β = 90°
(∵ tan (90 – θ) = cot θ)
So from given
tan 2A = cot (A – 27)
⇒ 2A + A – 27 = 90
⇒ 2A + A = 90 + 27
⇒ 3 A = 117 ⇒ A = \(\frac{117}{3}\) = 39
∴ A = 39°

Question 19.
Prove that
\(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Solution:

Question 20.
Evaluate (sin x – cos x)² + (sin x + cos x)².
Solution:
(sin x – cos x)² + (sin x + cos x)²
= sin² x + cos² x – 2 sin x cos x + sin² x + cos² x + 2 sin x cos x
= 2(sin²x + cos²x) = 2(1) = 2

Question 21.
cos A = \(\frac{12}{13}\), then, find sin A and tan A
Solution:
Given, cos A = \(\frac{12}{13}\)
AC² = AB² + BC²
132 = 12² + x²2
169 = 144 + x²
x² = 25 ⇒ x = 5

Question 22.
If 4 sin² θ – 1 = 0 then find θ’ (θ < 90) also, find the value of θ and the value of cos² θ + tan² θ
Solution:
Given 4 ² θ – 1 = 0 ⇒ 4 ² θ = 1
² θ = \(\frac{1}{4}\) ⇒ sin θ = ± \(\sqrt{\frac{1}{4}=} \pm \frac{1}{2}\)
Given θ is less than 90°
sin θ = \(\frac{1}{2}\)
sin θ = sin30° ⇒ θ = 30°
cos θ = cos = 30° \(\frac{\sqrt{3}}{2}\)
tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)
cos² θ + tan² θ = cos² 30° + tan² 30°
= (\(\frac{\sqrt{3}}{2}\))² + (\(\frac{1}{\sqrt{3}}\))²
= \(\frac{3}{4}+\frac{1}{3}\) = \(\frac{9+4}{12}\) = \(\frac{13}{12}\)

Question 23.
Prove that: (sin θ – cosec θ)² + (cos θ – sec θ)² = cot² θ + tan² θ – 1
Solution:
(sin θ – cosec θ)² + (cos θ – sec θ)²
= sin²θ + cosec²θ – 2 sinθ . cosecθ . + cos²θ + sec²θ – 2 cosθ . sec θ
= (sin²θ + cos²θ) + cosec²θ + sec²θ – 2 – 2
= 1 + (1 + cot²θ) + (1 + tan²θ) – 2 – 2
= cot²θ + tan²θ + 3 – 4
= cot²θ + tan²θ – 1

Question 24.
If cosec θ + cot θ = p, show that \(\frac{p^{2}+1}{p^{2}-1}\) = sec θ.
Solution:
cosec θ – cot θ = p
∴ cosec θ – cot θ = \(\frac{1}{\mathrm{p}}\) ……………….(1)
cosec θ + cot θ = p …………..(2)
(1) + (2), We get
2 cosec θ = p + \(\frac{1}{\mathrm{p}}\) = \(\frac{p^{2}+1}{p}\) …………..(3)
cosec θ – cot θ = p
cosec θ – cot θ = \(\frac{1}{\mathrm{p}}\)
(1) – (2), We get

Hence proved.

Question 25.
Prove that (1 + tan²θ) + (1 + \(\frac{1}{\tan ^{2} \theta}\)) = \(\frac{1}{\sin ^{2} \theta-\sin ^{4} \theta}\)
Solution:

Question 26.
If sec θ + tan θ = p, then prove that sin θ = \(\frac{\mathbf{p}^{2}-1}{\mathbf{p}^{2}+1}\)
Solution:
sec θ + tan θ = p
sec2 θ – tan2 θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
p. (sec θ – tan θ) = 1

Question 27.
If cot θ = \(\frac{7}{8}\) then,
Evaluate :
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
ii) \(\frac{1+\cos \theta}{\sin \theta}\)
Solution:

Question 28.
Show that
sec² θ + cosec² θ = sec² θ . cosec² θ.
Solution:
sec² θ + cosec² θ