# AP 10th Class Maths Important Questions Chapter 11 Trigonometry

These AP 10th Class Maths Chapter Wise Important Questions Chapter 11 Trigonometry will help students prepare well for the exams.

## AP State Syllabus 10th Class Maths 11th Lesson Important Questions and Answers 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