AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

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AP State Syllabus 10th Class Maths Bits 11th Lesson Trigonometry with Answers

Question 1.
Find the value of cos 12° – sin 78°
Answer:
0
Explanation:
cos 12°- sin(90°- 12°)
⇒ cos 12° – cos 12° = 0

Question 2.
If x = cosec θ + cot θ and y = cosec θ – cot θ, then write the relation between ‘x’ and ‘y’
Answer:
xy = 1
Explanation:
xy = (cosec θ + cot θ) (cosec θ – cot θ)
⇒ cosec² θ – cot² θ = 1

Question 3.
Write a formula to cos (A – B).
Answer:
cos A cos B + sin A sin B

Question 4.
The value of cos (90 – θ).
Answer:
sin θ

Question 5.
In Δ ABC sin C = \(\frac {3}{5}\),then find cos A.
Answer:
\(\frac {3}{5}\)

Question 6.
Complete the value tan² θ – sec² θ.
Answer:
-1
Explanation:
-(sec² θ – tan² θ) = – 1

Question 7.
The value of sec (90 – A).
Answer:
cosec A

Question 8.
If cosec θ + cot θ = 5, then cosec θ – cot θ.
Answer:
\(\frac {1}{5}\)

Question 9.
If x = 2 sec θ; y = tan θ,then the value of x² – y².
Answer:
4
Explanation:
sec² θ = \(\left(\frac{x}{2}\right)^{2}\),tan² θ = \(\left(\frac{y}{2}\right)^{2}\)
⇒ sec² θ – tan² θ = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)
⇒ 4 = x² – y²

Question 10.
If √3 tan θ = 1,then the value of θ.
A.
30°

Question 11.
The value of (sec 60)(cos 60).
Answer:
1

Question 12.
How much the value of sin (60 + 30)?
Answer:
1

Question 13.
If sec θ + tan θ = \(\frac{1}{2}\) then find sec θ – tan θ.
Answer:
2

Question 14.
The value of cos(90 – θ).
Answer:
sin θ

Question 15.
Simplify: tan 26°. tan 64°
Answer:
1
Explanation:
tan 26° . tan 64°
⇒ tan 26° . tan (90° – 26°)
⇒ tan 26° . cot 26° = 1

Question 16.
How much value of the angle ‘θ’ in the figure?

Answer:
30°
Explanation:
sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2}{4}=\frac{1}{2}\) = sin 30°
∴θ = 30°

Question 17.
Find the value of \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\).
Answer:
0
Explanation:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-1}{1+1}=\frac{0}{2}\) = 0

Question 18.
If sin x = \(\frac{5}{7}\), then find the value of cosec x.
Answer:
\(\frac{7}{5}\)
Explanation:
sin x = \(\frac{5}{7}\) ⇒ cosec x = \(\frac{7}{5}\)

Question 19.
Given ∠A = 75°, ∠B = 30°,then find the value of tan (A – B).
Answer:
1
Explanation:
tan (75° – 30°) = tan 45° = 1

Question 20.
If sec θ + tan θ = \(\frac{1}{3}\),then find the value of sec θ – tan θ.
Answer:
3
Explanation:
⇒ sec θ + tan θ = \(\frac{1}{3}\)
⇒ sec θ – tan θ = 3

Question 21.
If cosec θ + cot θ = 2,then find the value of cos θ.
Answer:
\(\frac{3}{5}\)
Explanation:

Question 22.
If cos (A + B) = 0, cos B = \(\frac{\sqrt{3}}{2}\),then
how much value of A?
Answer:
60°
Explanation:
cos (A + B) = 0, cos B = \(\frac{\sqrt{3}}{2}\)
⇒ cos B = cos 30°
⇒ A + B = 90°
⇒ A + 30° = 90°
∴ A = 90°- 30° = 60°

Question 23.
If sec θ – tan θ = 3, then find the value of sec θ + tan θ.
Answer:
\(\frac{1}{3}\)

Question 24.
If sin 2θ = cos 3θ, then how the value of θ.
Answer:
18°
Explanation:
sin 2θ = cos 3θ
⇒ 2θ + 3θ = 90°
⇒ 5θ = 90°
θ = 18°

Question 25.
If cos θ = \(\frac{3}{5}\), then find the value of sin θ.
Answer:
\(\frac{4}{5}\)
Explanation:
cos θ = \(\frac{3}{5}\) ⇒ sin θ = \(\frac{4}{5}\)

Question 26.
Simplify : cos 60° + sin 30°.
Answer:
1
Explanation:
cos 60° + sin 30° = \(\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\) = 1

Question 27.
If sec A + tan A = \(\frac{1}{5}\), then find the value of sec A – tan A.
Answer:
5

Question 28.
sin (90 – A) = \(\frac{1}{2}\) , then how much the value A?
A.
60°

Question 29.
If cot A = \(\frac{5}{12}\), then find the value of sin A + cos A.
Answer:
\(\frac{17}{13}\)

Question 30.
Write any value which is not possible for sin x?
Answer:
\(\frac{5}{4}\)
Explanation:
sin x = \(\frac{5}{4}\) > 1, so it is not possible.

Question 31.
If sin θ = cos θ (0 < θ < 90), then the value of tan θ + cot θ.
Answer:
2
Explanation:
sin θ = cos θ
⇒ 0 = 45°
⇒ tan 45° + cot 45° = 2

Question 32.
If sec θ + tan θ = 3, then find the value of sec θ – tan θ.
Answer:
\(\frac{1}{3}\)

Question 33.
In ΔABC; AB = c, BC = a, AC = b and ∠BAC = 0, then calculate the value of area of ΔABC is ………………(θ is acute)
Answer:
\(\frac{1}{2}\) bc sin θ

Question 34.
Write the value of tan θ in terms of cosec θ.
Answer:
\(\frac{1}{\sqrt{\operatorname{cosec}^{2} \theta-1}}\)

Question 35.
Observe the following:
(I) sin² 20° + sin² 70° = 1
(II) log₂ (sin 90°) = 1
Which one is CORRECT?
Answer:
(I) only

Question 36.
Simplify : tan 36°. tan 54° + sin 30°
Answer:
\(\frac{3}{2}\)
Explanation:
tan 36° . tan (90 – 36°) + sin 30°
⇒ tan 36° . cot 36° + \(\frac{1}{2}\)
⇒ 1 + \(\frac{1}{2}=\frac{3}{2}\)

Question 37.
If sin A = \(\frac{24}{25}\), then find the value of sec A.
Answer:
\(\frac{25}{7}\)

Question 38.
Which one of the following is NOT defined?
sin 90°, cos 0°, sec 90°, cos 90°.
Answer:
sec 90°
Explanation:
sec 90° is not defined

Question 39.
Simplify:\(\sqrt{\frac{1-\cos ^{2} A}{1+\cot ^{2} A}}\)
Answer:
sin²A
Explanation:
\(\sqrt{\frac{\sin ^{2} A}{\operatorname{cosec}^{2} A}}\) = \(\sqrt{\sin ^{4} A}\) = sin²A

Question 40.
If cot θ – cosec θ = p, then find cot θ + cosec θ.
Answer:
\(\frac{-1}{\mathrm{p}}\)

Question 41.
Express tan θ in terms of cos θ.
Answer:
\(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\)

Question 42.
Who was introduced Trigonometry?
Answer:
Hipparchus

Question 43.
\(\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}\) equal to
Answer:
1
Explanation:

Question 44.
sin²47° + sin²43° equal to
Answer:
1

Question 45.
sin 2A equal to
Answer:
2sin A cos A

Question 46.
sin 30° + cos 60° equal to
Answer:
1

Question 47.
sec⁴A – sec²A equal to
Answer:
tan⁴ A – tan² A
Explanation:
sec⁴ A – sec²A
⇒1 + tan⁴ A – (1 + tan² A)
⇒ tan⁴ A – tan² A

Question 48.
Find the value of \(\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}\).
Answer:
sin² θ

Question 49.
2sin 45°. cos 45° equal to
Answer:
1

Question 50.
sin 81° equal to
Answer:
cos 9°
1

Question 51.
tan θ = \(\frac{1}{\sqrt{3}}\), then find cos θ.
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 52.
If cos θ = \(\frac{1}{2}\); then find cos\(\frac{\theta}{2}\)
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
cos θ = \(\frac{1}{2}\)
⇒ cos θ = cos 60° ⇒ θ = 60°
⇒ cos\(\frac{\theta}{2}\) ⇒ cos 30° = \(\frac{\sqrt{3}}{2}\)

Question 53.
\(\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}\) equal to
Answer:
cos θ + sin θ

Question 54.
cos 300° equal to
Answer:
\(\frac{1}{2}\)
Explanation:
cos (270° + 30°) = sin 30° =\(\frac{1}{2}\)

Question 55.
\(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\) equal to
Answer:
tan θ
Explanation:

Question 56.
(1 + tan² θ)cos² θ equal to
Answer:
1

Question 57.
If 3 tan θ = 1; then find θ.
Answer:
30°

Question 58.
\(\frac{\sqrt{\sec ^{2} \theta-1}}{\sec \theta}\) equal to
Answer:
sin θ
Explanation:

Question 59.
\(\frac{1}{\sqrt{1+\tan ^{2} \theta}}\) equal to
Answer:
cos θ

Question 60.
Find the value of cosec 60° × cos 90°.
Answer:
0

Question 61.
sec² θ + cosec² θ equal to
Answer:
sec² θ.cosec² θ
Explanation:

Question 62.
sec² 33° – cot² 57° equal to
Answer:
1

Question 63.
If sin θ = \(\frac{11}{15}\), then find cos θ.
Answer:
\(\frac{2 \sqrt{26}}{15}\)

Question 64.
equal to \(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\)
Answer:
sin θ

Question 65.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\); cos θ = \(\frac{\mathbf{c}}{\mathbf{d}}\); then find tan θ.
Answer:
\(\frac{\mathrm{ad}}{\mathrm{bc}}\)
Explanation:

Question 66.
tan (A+B) equal to
Answer:
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\)

Question 67.
If sin A = \(\frac{1}{\sqrt{2}}\) ; then find tan A.
Answer:
1

Question 68.
sin\(\frac{\pi}{6}\) + cos\(\frac{\pi}{3}\) equal to
Answer:
1

Question 69.
Find the value of cos 75°.
Answer:
sin 15°
Explanation:
cos 75° = cos (90° – 15°) = sin 15°

Question 70.
If sin 0 = \(\frac{1}{2}\); then find cos \(\frac{3 \theta}{2}\).
Answer:
\(\frac{1}{\sqrt{2}}\)

Question 71.
\(\frac{1}{\sec ^{2} A}+\frac{1}{\operatorname{cosec}^{2} A}\) equal to
Answer:
1
Explanation:

Question 72.
sin 240° equals to
Answer:
–\(\frac{\sqrt{3}}{2}\)
Explanation:
sin 240° = sin (270° – 30°)
= -cos 30°= \(\frac{-\sqrt{3}}{2}\)

Question 73.
If sec θ + tan θ = \(\frac{1}{5}\) ,then find sin θ.
Answer:
\(\frac{12}{13}\)

Question 74.
tan 240° equal to
Answer:
√3

Question 75.
Find the value of
sin 60° cos 30°+ cos 60°. sin 30°.
Answer:
1
Explanation:

Question 76.
\(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) equal to
Answer:
2sec² θ
Explanation:

Question 77.
tan 0° equal to
Answer:
0

Question 78.
\(\frac{\sqrt{1+\tan ^{2} \theta}}{\sqrt{1+\cot ^{2} \theta}}\) equal to
Answer:
tan θ

Question 79.
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) equal to
Answer:
1

Question 80.
If sin θ = cos θ, then find θ.
Answer:
45°

Question 81.
In right triangle ΔABC; ∠B= 90°; tan C = \(\frac{5}{12}\) , then find the length of hypotenuse.
Answer:
13
Explanation:

By Py thagoras theorem,
AC² = AB² + BC²
AC² = 5² + 12²
= 25 + 144= 169
⇒ AC = hypotenuse = 13

Question 82.
If A, B are acute angles ;
sin(A – B)= \(\frac{1}{2}\); cos (A + B) = \(\frac{1}{2}\), then find B.
Answer:
15° (or) \(\frac{\pi}{12}\)
Explanation:
sin (A – B) = \(\frac{1}{2}\) = sin30°
A – B = 30°

Question 83.
cos (270° – θ) equal to
Answer:
-sin θ

Question 84.
Find the value of
cos 0° + sin 90° + √2sin 45°.
Answer:
3
Explanation:
1 + 1 + √2.\(\frac{1}{\sqrt{2}}\) = 1 + 1 + 1 = 3

Question 85.
If tan θ = \(\frac{1}{\sqrt{3}}\); then find cos θ.
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
⇒ θ = 30°
∴ cos θ = cos 30° = \(\frac{\sqrt{3}}{2}\)

Question 86.
Equal to cosec (90 + θ).
Answer:
sec θ

Question 87.
sin θ. sec θ equals to
Answer:
tan θ

Question 88.
Find the value of 3sin² 45°+2cos² 60°.
Answer:
2
Explanation:

Question 89.
Find the value of tan² 30° + 2 cot² 60°.
Answer:
1
Explanation:

Question 90.
Find the value of secA.\(\sqrt{1-\sin ^{2} A}\)
Answer:
1

Question 91.
If 5 sin A= 3; then find sec² A – tan² A.
Answer:
1
Explanation:
sin A = \(\frac{3}{5}\) ⇒ sec²A = tan²A = 1

Question 92.
Find the value of cos 240°.
Answer:
–\(\frac{1}{2}\)

Question 93.
If sin θ. cosec θ = x; then find x.
Answer:
1

Question 94.
sin(45°+ θ) – cos(45°- θ).
Answer:
0

Question 95.
Find the value of cos² 17° – sin² 73°.
Answer:
0
Explanation:
cos² 17°- sin² 73°
= cos² (90 – 73) – sin² 73°
= sin² 73° – sin² 73° = 0

Question 96.
Find the value of sin² 60° – sin² 30°.
Answer:
\(\frac{1}{2}\)

Question 97.
ten θ is not defined when θ is equal to
Answer:
90°

Question 98.
Find the value of sin 45° + cos 45°.
Answer:
√2
Explanation:

Question 99.
Simplify: \(\frac{1-\sec ^{2} A}{\operatorname{cosec}^{2} A-1}\)
Answer:
– tan⁴A

Question 100.
Find the value of sin θ. cosec θ+ cos θ. sec θ + tan θ. cot θ.
Answer:
3
Explanation:

= 1 + 1 + 1 = 3

Question 101.
If A = 30°, then sin 2A equals to
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 102.
If sec θ = 3k and tan θ = \(\frac{3}{\mathbf{k}}\), then find the value of (k² – \(\frac{1}{\mathbf{k}^{2}}\))
Answer:
\(\frac{1}{9}\)
Explanation:

Question 103.
If tan θ + sec θ = 8, then find sec θ – tan θ.
Answer:
\(\frac{1}{8}\)

Question 104.
If sin θ = \(\frac{12}{13}\), then find tan θ.
Answer:
\(\frac{12}{5}\)

Question 105.
In a right angled ΔABC, right angle at C if tan A = \(\frac{8}{15}\), then find the value of cosec² A – 1.
Answer:
\(\frac{225}{64}\)
Explanation:
tan A = \(\frac{8}{15}\),
cosec ² A – 1 = cot²A = \(\left(\frac{15}{8}\right)^{2}=\frac{225}{64}\)

Question 106.
Find the value of \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)
Answer:
sin 60°

Question 107.
If sin A = ;\(\frac{1}{\sqrt{2}}\) then find tan A
Answer:
1

Question 108.
In ΔABC, sin \(\left(\frac{B+C}{2}\right)\) equal to .
Answer:
A. cos\(\frac{A}{2}\)

Question 109.
tan 26°. tan 64° equal to
Answer:
1
Explanation:
tan 26° . tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° = 1

Question 110.
cos² θ equal to
Answer:
1 – sin² θ

Question 111.
If tan A = \(\frac{3}{4}\) then find sec² A – tan² A.
Answer:
1

Question 112.
sin⁴ θ – cos⁴ θ equal to
Answer:
2sec² θ – 1

Question 113.
\(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\) equal to
Answer:
tan θ
Explanation:

Question 114.
sin(90 – Φ) equal to
Answer:
cos Φ

Question 115.
sec θ – tan θ = \(\frac{1}{n}\), then find sec θ + tan θ.
Answer:
n

Question 116.
x = 2 cosec θ, y = 2 cot θ; find x² – y².
Answer:
4

Question 117.
tan θ is not defined if θ.
Answer:
90°

Question 118.
sec θ is not defined if θ.
Answer:
90°

Question 119.
tan² Φ – sec² Φ equal to
Answer:
-1

Question 120.
\(\left|\begin{array}{ll}
\tan \theta & \sec \theta \\
\sec \theta & \tan \theta
\end{array}\right|\)
Answer:
1
Explanation:
\(\left|\tan ^{2} \theta-\sec ^{2} \theta\right|=|-1|\) = 1

Question 121.
sin 225° equal to
Answer:
\(\frac{-1}{\sqrt{2}}\)

Question 122.
cos (x – y) equal to
Answer:
cos x cos y + sin x sin y

Question 123.
\(\frac{\sec 35^{\circ}}{\operatorname{cosec} 55^{\circ}}\) equal to
Answer:
1
Explanation:

Question 124.
\(\frac{1}{\sec ^{2} A}+\frac{1}{\operatorname{cosec}^{2} A}\) equal to
Answer:
1

Question 125.
sin (-θ) equal to
Answer:
-sin θ

Question 126.
cosec (270 – θ) equal to
Answer:
-sec θ

Question 127.
sec (90 + θ) equal to
Answer:
-cosec θ

Question 128.
tan (360 – θ) equal to
Answer:
-tan θ

Question 129.
cos (-θ) equal to
Answer:
cos θ

Question 130.
sin (180 – θ) equal to
Answer:
sin θ

Question 131.
cos (270 – θ) equal to
Answer:
-sin θ

Question 132.
Find the maximum value of cos θ.
Answer:
1

Question 133.
Find the minimum and maximum values of tan θ.
Answer:
(- ∞ ,∞ )

Question 134.
sin 420° equal to
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 135.
sec 240° equal to
Answer:
-2

Question 136.
cos 0° + sin 90° + √3 cosec 60° equal to
Answer:
4
Explanation:
1 + 1 + 3 . \(\frac{2}{\sqrt{3}}\) = 4

Question 137.
sec θ + tan θ = \(\frac{1}{2}\); then find sin θ.
Answer:
\(\frac{12}{13}\)

Question 138.
sin 45°.cos 45° + √3 sin 60° equal to
Answer:
2

Question 139.
tan 30° + cot 30° equal to
Answer:
\(\frac{4}{\sqrt{3}}\)

Question 140.
tan (A – B) equal to
Answer:
\(\frac{\tan A-\tan B}{1+\tan A \tan B}\)

Question 141.
If sin A = \(\frac{3}{5}\); then find sin (90 + A).
Answer:
\(\frac{4}{5}\)
Explanation:
sin A = \(\frac{3}{5}\), sin (90 + A) = cos A = \(\frac{4}{5}\)

Question 142.
Find the value of cos 1°.cos 2°.cos 3°……………, cos 180°.
Answer:
0
Explanation:
cos 1° × cos 2° × cos 3° × …………….. × cos 90° × ……….. × cos 180°
cos 1° × cos 2° × cos 3° × …………. × 0 × …………. × (- 1) = 0

Question 143.
Find the value of cos²17° – sin² 73°.
Answer:
0

Question 144.
If cosec θ + cot θ = 2; then find cosec θ – cot θ.
Answer:
\(\frac{1}{2}\)

Question 145.
cosec 60°. sec 60° equal to
Answer:
\(\frac{4}{\sqrt{3}}\)

Question 146.
sin (A – B) equal to
Answer:
sin A cos B – cos A sin B

Question 147.
If tan (15°+ B)= √3 ; then find B.
Answer:
45°
Explanation:
tan (15° + B) = √3 = tan 60°
15 + B = 60 ⇒B = 60°- 15° = 45°

Question 148.
Simplify: \(\frac{\sin (90-\theta) \sin \theta}{\tan \theta}\) – 1
Answer:
-sin² θ

Question 149.
sin 450° equal to
Answer:
1

Question 150.
cos 150° equal to
Answer:
–\(\frac{\sqrt{3}}{2}\)

Question 151.
If sin θ = \(\frac{1}{2}\) ; then find cot θ.
Answer:
√3

Question 152.
Find the value of sin 29° – cos 61°
Answer:
0

Question 153.
If tan θ = 1; then find cos θ.
Answer:
\(\frac{1}{\sqrt{2}}\)

Question 154.
cos (A + B) equal to
Answer:
cos A cos B – sin A sin B

Question 155.
Express tan θ, in terms of sin θ.
Answer:
\(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\)

Question 156.
sin θ + sin² θ = 1, then find cos² θ + cos⁴ θ.
Answer:
1

Question 157.
in ΔABC,write tan(\(\frac{B+C}{2}\))equal to
Answer:
Cot (\(\frac{\mathrm{A}}{2},\))
Explanation:
A + B + C = 180°

Question 158.
(cos A + sin A)² + (cos A – sin A)² equal to
Answer:
2

Question 159.
cos(180 – θ) equal to
Answer:
– cos θ

Question 160.
Find the value of (cosec θ – cot θ).
Answer:
\(\frac{1-\cos \theta}{\sin \theta}\)
Explanation:
cosec θ – cot θ
= \(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\frac{1-\cos \theta}{\sin \theta}\)

Question 161.
sin² 75° + cos² 75° equal to
Answer:
1

Question 162.
If cos θ. sin θ = \(\frac{1}{2}\) ; then find tan θ.
Answer:
1
Explanation:
cos θ . sin θ = cos 45° . sin 45°
= \(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}\)
Then tan 45° = 1.

Question 163.
For which value of cosine equal to sin 81°.
Answer:
cos 9°.

Question 164.
sin 750° equal to
Answer:
\(\frac{1}{2}\)

Question 165.
\(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) equal to
Answer:
tan θ
Explanation:

Question 166.
If sin(A+B) = 1; sin B = \(\frac{1}{2}\) ; then find A.
Answer:
60°
Explanation:
sin (A + B) = 1 = sin 90°
A + B = 90°
sin B = sin 30° ⇒ B = 30°
A + 30° = 90° ⇒ A = 60°

Question 167.
If tan θ = √3 , then find sec θ.
Answer:
2
Explanation:
tan θ = √3 = tan 60° ⇒ θ = 60°
sec 60° = 2

Question 168.
Find the value of
cos 0°+ sin 90° + √3 cosec 60°.
Answer:
4

Question 169.
\(\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}\) equal to
Answer:
sec θ . cosec θ
Explanation:

Question 170.
Find the value of
cos 60°. cos 30° – sin 60°. sin30°.
Answer:
0

Question 171.
cot²θ – \(\frac{1}{\sin ^{2} \theta}\) equal to
Answer:
-1
Explanation:
cot² θ – cosec² θ = – 1

Question 172.
π radians equal into degrees.
Answer:
180°

Question 173.
sin (A + B). cos(A – B) + sin (A – B). cos (A + B) equal to
Answer:
sin 2A

Question 174.
If cos (A+B) = 0, cos B = \(\frac{\sqrt{3}}{2}\) ;then find A.
Answer:
60°

Question 175.
cos⁶ θ + sin⁶ θ equal to .
Answer:
1 – 3 sin² θ.cos² θ
Explanation:
cos⁶ θ + sin⁶ θ = (cos² θ)³ + (sin² θ)³
a³ + b³ = (a + b) – 3ab (a + b)
= (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin² θ + cos² θ)
= 1 – 3 sin² θ cos² θ (1)
= 1 – 3 sin² θ cos² θ

Question 176.
sin 225° equal to
Answer:
–\(\frac{1}{\sqrt{2}}\)

Question 177.
sin 180° equal to
Answer:
0

Question 178.
If x = 2 cosec θ; y = 2 cot θ; then find x² – y².
Answer:
4
Explanation:
\(\frac{\mathrm{x}}{2}\) = cosec θ, \(\frac{\mathrm{y}}{2}\) = cot θ
cosec² θ – cot² θ = \(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2}\right)^{2}\)
1 = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)
x² – y² = 4

Question 179.
cos θ. tan θ equal to
Answer:
sin θ
Explanation:
cos θ = \(\frac{\sin \theta}{\cos \theta}\) = sin θ

Question 180.
If cot² θ = 3; then find cosec θ.
Answer:
2
Explanation:
cot θ = √3 = cot 30° ⇒ θ = 30°
∴ cosec 30° = 2.

Question 181.
If sec θ = cosec θ; then find the value of θ.
Answer:
\(\frac{\pi}{4}\)

Question 182.
\(\frac{\tan \theta \cdot \sqrt{1-\sin ^{2} \theta}}{\sqrt{1-\cos ^{2} \theta}}\) equal to
Answer:
1
Explanation:

Question 183.
cos(x – y) equal to
Answer:
cos x cos y + sin x sin y

Question 184.
Simplify : cosec 31° – sec 59°
Answer:
0
Explanation:
cosec 31° – sec (90° – 31°)
[∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31°
= 0

Question 185.
(sec 45° + tan 45°) (sec 45° – tan 45°) equal to
Answer:
1

Question 186.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\); cos θ = \(\frac{\mathbf{c}}{\mathbf{d}}\), then find cot θ.
Answer:
\(\frac{\mathrm{bc}}{\mathrm{ad}}\)

Question 187.
\(\sqrt{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}\) equal to
Answer:
1

Question 188.
sin² 45° + cos² 45° + tan² 45° equal to
Answer:
2
Explanation:

Question 189.
sec (360° – θ) equal to
Answer:
sec θ

Question 190.
tan θ. cot θ = sec θ. x ; then find x.
Answer:
cos θ
Explanation:
tan θ . cot θ = sec θ . x
\(\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}=\frac{x}{\cos \theta}\)
cos θ = x

Question 191.
If 4 sin 30°. sec 60° = x tan 45°; then find x.
Answer:
4
Explanation:
4 . sin 30° – sec 60° = x . tan 45°
4 . \(\frac{1}{2}\) . 2 = x . 1
⇒ x = 4

Question 192.
In the following which are in geometric progression?
A) sin 30°, sin 45°, sin 60°
B) sec 30°, sec 45°, sec 60°
C) tan 30°, tan 45°, tan 60°
D) cos 45°, cos 60°, cos 90°
Answer:
C) tan 30°, tan 45°, tan 60°

Question 193.
(1 + tan² A) (1 – sin² A)
equal to
Answer:
1
Explanation:
sec² A × cos² A = 1

Question 194.
Find the value of
cos 60° cos 30° + sin 60°. sin 30°.
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
\(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}\)

Question 195.
\(\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) equal to
Answer:
\(\frac{1}{2}\)

Question 196.
cos(\(\frac{3 \pi}{2}\) + θ) equal to
Answer:
sin θ
Explanation:
cos (270 + θ) = sin θ

Question 197.
\(\sqrt{1+\cot ^{2} \theta}\) equal to
Answer:
cosec θ

Question 198.
If tan θ + cot θ = 2; then find tan² θ + cot² θ.
Answer:
2
Explanation:
tan θ + cot θ = 2
⇒ tan² θ + cot² θ + 2 . tan θ . cot θ = 4
⇒ tan² θ + cot² θ = 4 – 2 = 2

Question 199.
If sec θ = \(\frac{13}{12}\), then find sin θ.
Answer:
\(\frac{5}{13}\)

Question 200.
How much the value of
(sin θ + cos θ)² + (sin θ – cos θ)² ?
Answer:
2
Explanation:
(a + b)² + (a – b)² = 2(a² + b²)
= 2 (sin² θ + cos² θ) = 2

Question 201.
(1 + tan θ)² equal to
Answer:
sec² θ + 2 tan θ

Question 202.
sin(A – B) = \(\frac{1}{2}\); cos (A+B) = \(\frac{1}{2}\). So
find A.
Answer:
45°
Explanation:

⇒ A = 45°

Question 203.
Find the value of tan 135°.
Answer:
-1

Question 204.
Find the value of \(\sqrt{1+\sin A} \cdot \sqrt{1-\sin A}\)
Answer:
cos A
Explanation:

Question 205.
Find the value of tan 60° – tan 30°.
Answer:
\(\frac{2 \sqrt{3}}{3}\)
Explanation:
tan 60° – tan 30°

Question 206.
In ΔABC, a = 3; b = 4; c = 5, then find cos A.
Answer:
4/5
Explanation:

Question 207.
sin³ θ cos θ.cos³ θ sin θ equals to
Answer:
sin θ cos θ

Question 208.
If tan θ \(\frac{1}{\sqrt{3}}\) , then find the value of 7 sin² θ + 3 cos² θ.
Answer:
4
Explanation:

Question 209.
cot (270° – θ) equal to
Answer:
tan θ

Question 210.
(1 + tan² 60°)² equals to
Answer:
16
Explanation:
[1 + (√3)²]² = (1 + 3)² = 4² = 16

Question 211.
sin (270° + θ) equal to
Answer:
– cos θ

Question 212.
If tan² 60° + 2 tan² 45° = x tan 45°, then find x.
Answer:
5
Explanation:
(√3)² + 2(1)² = x . 1
⇒ 3 + 2 = x ⇒ x = 5

Question 213.
cos² 0° + cos² 60° equal to
Answer:
5/4

Question 214.
Simplify : sin⁴ θ – cos⁴ θ
Answer:
2sin² θ – 1

Question 215.
If α + β = 90° and α = 2β, then find cos² β + sin² β.
Answer:
1
Explanation:
α = 90 – β
⇒ 90 – β = 2β ⇒ 3β = 90° ⇒ β = 30°
∴ cos² 30° + sin² 30° = (\(\frac{\sqrt{3}}{2}\))² + (\(\frac{1}{2}\))²
= \(\frac{3}{4}+\frac{1}{4}=\frac{4}{4}\) = 1

Question 216.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\) , then find cos θ.
Answer:
\(\frac{\sqrt{b^{2}-a^{2}}}{b}\)

Question 217.
2sin θ = sin² θ is true for the value of θ is
Answer:
0°

Question 218.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\) , then find tan θ.
Answer:
\(\frac{a}{\sqrt{b^{2}-a^{2}}}\)

Question 219.
\(\frac{\sin \theta}{1+\cos \theta}\) is equal to
Answer:
\(\frac{1-\cos \theta}{\sin \theta}\)

Question 220.
If sin θ = cos θ, then find the value of 2 tan θ + cos² θ.
Answer:
\(\frac{5}{2}\)
Explanation:
sin θ = cos θ ⇒ θ = 45°
2 tan 45° + cos² 45° = 2 + \(\frac{1}{2}=\frac{5}{2}\)

Question 221.
If sin x = cos x, 0 ≤ x ≤ 90°, then find x.
Answer:
45°

Question 222.
How much the maximum value of sin θ?
Answer:
1

Question 223.
In the figure find tan x.

Answer:
\(\frac{15}{8}\)

Question 224.
sin A = cos B, then find A + B.
Answer:
90°

Question 225.
If cosec θ + cot θ = 3, then find cosec θ – cot θ.
Answer:
\(\frac{1}{3}\)

Question 226.
If tan 2A = cot (A – 18°) where 2A is an acute angle, then find A.
Answer:
36°
Explanation:
90 – 2A = A – 18°
⇒ 3A = 108° ⇒ A = 36°

Question 227.
sec 0° equal to
Answer:
1

Question 228.
cosec 300° equal to
Answer:
\(\frac{-2}{\sqrt{3}}\)

Question 229.
cos 240° equal to
Answer:
\(\frac{-1}{2}\)

Question 230.
\(\frac{\operatorname{cosec}^{2} \theta}{\cot \theta}\) – cot θ equal to
Answer:
tan θ

Question 231.
Find the minimum value of cos θ.
Answer:
-3

Question 232.
If sec θ = cosec θ; then find the value of θ in radians.
Answer:
\(\frac{\pi^{\mathrm{c}}}{4}\)
Explanation:
sec θ = cosec θ ⇒ θ = 45° = \(\frac{\pi^{\mathrm{c}}}{4}\)

Question 233.
Reciprocal of cot A.
Answer:
tan A

Question 234.
sin \(\frac{\pi^{\mathrm{c}}}{4}\) + cos 45° equal to
Answer:
√2

Question 235.

Answer:
\(\frac{\mathrm{x}}{\mathrm{z}}\)

Question 236.
If cosec θ = \(\frac{25}{7}\), then find cot θ.
Answer:
\(\frac{24}{7}\)

Question 237.
If sin (A + B) = \(\frac{\sqrt{3}}{2}\); cos B = \(\frac{\sqrt{3}}{2}\) , then find A.
Answer:
30°

Question 238.
tan 750° equal to
Answer:
\(\frac{1}{\sqrt{3}}\)
Question 239.
(1 – sec² θ) (1 – cosec² θ) equal to
Answer:
1

Question 240.
If cosec θ – cot θ = 4, then find cosec θ + cot θ.
Answer:
\(\frac{1}{4}\)

Question 241.
\(\sqrt{\tan ^{2} \theta+\cot ^{2} \theta+2}\) equal to
Answer:
tan θ + cot θ
Explanation:
= \(\sqrt{1+\tan ^{2} \theta+1+\cot ^{2} \theta}\)
= \(\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}\)
= tan θ + cot θ

Question 242.
If cos θ = \(\frac{3}{5}\); then cos (-θ) equal to
Answer:
\(\frac{3}{5}\)

Question 243.
\(\frac{\sqrt{\operatorname{cosec}^{2} \theta-1}}{\operatorname{cosec} \theta}\)
equal to
Answer:
cos θ

Question 244.
If 5 sin A = 3, then find sec² A – tan² A.
Answer:
1

Question 245.
In the figure, find AB.

Answer:
20√3 (or) \(\frac{60}{\sqrt{3}}\)

Question 246.
If cot θ = x; then find cosec θ.
Answer:
\(\sqrt{\mathrm{x}_{1}^{2}+1}\)
Explanation:
cot θ = x ⇒ cosec θ = \(\sqrt{x^{2}+1}\)

Question 247.
\(\sqrt{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta-\cos ^{2} \theta}\) equal to
Answer:
cot θ

Question 248.
Find the value of \(\frac{1}{\sec \theta-\tan \theta}\)
Answer:
sec θ + tan θ

Question 249.
sin (A + B) equal to
Answer:
sin A cos B + cos A sin B

Question 250.
\(\sqrt{(\sec \theta+1)(\sec \theta-1)}\) equal to
Answer:
tan θ

Question 251.
Find the value of tan 5° × tan 30° × 4 tan 85°.
Answer:
\(\frac{4}{\sqrt{3}}\)

Question 252.
cos 110°.cos 70° – sin 110°.sin 70° equal to ,
Answer:
-1
Explanation:
cos A . cos B – sin A . sin B
= cos (A + B)
= cos (110 + 70) = cos 180° = – 1

Question 253.
Find the value of tan 1°.tan 2°.tan 3°………….tan 89°.
Answer:
1

Question 254.
If cos θ = -cos θ; then Write θ in radian measure.
Answer:
\(\frac{\pi^{\mathrm{c}}}{3}\)

Question 255.
sec A = cosec B, then write A and B are ……….. angles.
Answer:
Complementary.

Question 256.
Find the value of tan 75°.
Answer:
2 + √3

Question 257.
\(\sqrt{\sec ^{2} \theta-\tan ^{2} \theta+\cot ^{2} \theta}\) equal to
Answer:
cosec θ

Question 258.
sin 240° + sin 120° equal to
Answer:
0

Question 259.
Find the value of
sec 70°. sin 20° + cos 20°. cosec 70°.
Answer:
2

Question 260.
If sec A + tan A = \(\frac{1}{3}\); then find sec A – tan A.
Answer:
3

Question 261.
(sec² θ – 1)(1 – cosec² θ)equal to
Answer:
-1

Question 262.
cos θ equal to
Answer:
\(\frac{\cot \theta}{\operatorname{cosec} \theta}\)

Question 263.
The radius of a circle is ‘r’; an arc of length ‘L’ is making an angle θ, at the centre of the circle, then find θ.
Answer:
L/r

Question 264.
cos (A – B) = \(\frac{1}{2}\); sin B = \(\frac{1}{\sqrt{2}}\), find measure of A.
Answer:
105°

Question 265.
If sec θ + tan θ = 4; then find cos θ.
Answer:
\(\frac{8}{17}\)

Question 266.
sec (360° – θ) equals to
Answer:
sec θ

Question 267.
If A is acute and tan A = \(\frac{1}{\sqrt{3}}\); then find sin ‘A’.
Answer:
\(\frac{1}{2}\)

Question 268.
(1 +cot² 45°)² equal to
Answer:
4

Question 269.
\(\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}\) equal to
Answer:
2\(\frac{1}{2}\)

Question 270.
\(\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}\) equal to
Answer:
sec x + tan x

Question 271.
\(\frac{\sin ^{4} A-\cos ^{4} A}{\sin ^{2} A-\cos ^{2} A}\) equal to
Answer:
1

Question 272.
If the angle in a triangle are in the ratio of 1:2:3, then find the smallest angle in radins.
Answer:
π/6

Question 273.
If sin θ + cos θ = √2; then find the value of ‘θ’.
Answer:
45°

Question 274.
If cosec θ = 2 and cot θ = √3 P; where θ is an acute angle, then find the value of ‘P’.
Answer:
1

Question 275.
If cos 2θ = sin 4θ; here 2θ and 4θ are acute angles, then find the value of θ.
Answer:
15°
Explanation:
2θ + 4θ = 90°
⇒ 6θ = 90°⇒ θ = \(\frac{90^{\circ}}{6}\) = 15°

Question 276.
If sin 45°.cos 45°+cos 60° = tan θ, then find the value of θ.
Answer:
45°

Question 277.
If P, Q and R are interior angles of a ΔPQR, then tan \(\left(\frac{\mathbf{P}+\mathbf{Q}}{2}\right)\) equals
Answer:
cot (\(\frac{\mathrm{R}}{2}\))

Question 278.
If tan θ = 1, then find the value of \(\frac{5 \sin \theta+4 \cos \theta}{5 \sin \theta-4 \cos \theta}\)
Answer:
9
Explanation:
tan θ = 1 = tan 45° ⇒ θ = 45°

Question 279.
If sec 2A = cosec (A – 27°), where 2A is an acute angle, then find the measure of ∠A.
Answer:
39°
Explanation:
90 – 2A = A – 27°
⇒ 117° = 3A ⇒ A = \(\frac{117^{\circ}}{3}\) = 39°

Question 280.
If sin C = \(\frac{3}{5}\); then find cos A.

Answer:
3/5

Question 281.
Expressing tan θ, interms of sec θ.
Answer:
\(\sqrt{\sec ^{2} \theta-1}\)

Question 282.
If sin θ. cos θ = k; then find sin θ + cos θ.
Answer:
\(\sqrt{1+2 \mathrm{k}}\)

Question 283.
If \(\frac{1}{2}\) tan² 45° = sin2 A and ’A’ is acute, then find the value of ‘A’.
Answer:
45°

Question 284.
Find the value of (\(\frac{11}{\cot ^{2} \theta}-\frac{11}{\cos ^{2} \theta}\))
Answer:
-11
Explanation:
11 (tan² θ – sec² θ) = 11 (-1) = -11

Question 285.
Find the maximum value of \(\frac{1}{\sec \theta}\)
0° ≤ θ ≤ 90°.
Answer:
1

Question 286.
If π < θ < \(\frac{3 \pi}{2}\), then θ lies in which quadrant?
Answer:
Third quadrant

Question 287.
If cos θ = \(\frac{\sqrt{3}}{2}\) and’θ’is acute, then find the value of 4sin² θ + tan² θ.
Answer:
4/3

Question 288.
If tan θ = \(\frac{7}{8}\), then find the value of \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) ,
Answer:
\(\frac{64}{49}\)

Question 289.
When 0° ≤ 0 ≤ 90°; find the maximum value of sin θ + cos θ.
Answer:
√2

Question 290.
In ΔABC, ∠B = 90° ; ∠C = θ. From the figure find tan θ.

Answer:
\(\frac{15}{8}\)

Question 291.
If sin (x – 20°) = cos(3x – 10°), then find x’.
Answer:
15°

Question 292.
If sin (A – B)= \(\frac{1}{2}\); cos (A + B)= \(\frac{1}{2}\),
then find ‘B’.
Answer:
15°

Question 293.
If 5 tan θ = 4, then find die value of
\(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta}\)
Answer:
\(\frac{1}{7}\)

Question 294.
If 4 cos2 θ – 3 = 0, then find the value of sin θ.
Answer:
\(\frac{1}{2}\)

Choose the correct answer satisfying the following statements.
Question 295.
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
Explanation:
sin² θ + cos² θ = 1
⇒ sin² 67° + cos² 67° = 1
Hence, (i) is the correct option.

Question 296.
Statement (A): If cos A + cos² A = 1,
then sin² A + sin⁴ A = 2
Statement (B): 1 – sin² A = cos² A, for any value of A.
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:
iii) A is false, B is true
Explanation:
cos A + cos² A = 1
cos A = 1 – cos² A = sin² A
∴ sin² A + sin⁴ A = cos A + cos² A = 1
⇒ sin² A + sin⁴ A = 1
Hence, (iii) is the correct option.

Question 297.
Statement (A):
The value of sec² 10° – cot2 80° is 1.
Statement (B):
The value of sin 30° = \(\frac{1}{2}\)
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
Explanation:
We have sec² 10° – cot² 80°
= sec² 10°-cot² (90°- 10°)
= sec² 10° – tan² 10° = 1
Also, sin 30° = \(\frac{1}{2}\).
Hence, (i) is the correct option.

Question 298.
Statement (A) : The value of sin θ cos (90 – θ) + cos θ sin(90 – θ) ‘ equal to 1.
Statement (B): tan θ = sec(90 – θ)
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:
ii) A is true, B is false
Explanation:
sin θ cos (90 – θ) + Cos θ sin (90 – θ)
= sin θ . sin θ + cos θ – cos θ
= sin² θ + cos² θ = 1 and tan θ = cot (90 – θ)
Hence, (ii) is the correct option.

Question 299.
Statement (A) : The value of sin θ = \(\frac{4}{3}\) is not possible.
Statement (B): Hypotenuse is the largest side in any right angled triangle.
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
Explanation:
sin ² = \(\frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{3}\)
Here, perpendicular is greater than the hypotenuse which is not possible in any right triangle.
Hence, (i) is the correct option.

Question 300.
Statement (A) : In a right angled triangle, if tan θ = \(\frac{3}{4}\), the greatest side of
the triangle is 5 units.
Statement (B) : (Greatest side hypotenuse)² = (perpendicular)² – (base)²
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
Explanation:
Both A and B are correct and B is the correct explanation of the A.
Greatest side = \(\sqrt{(3)^{2}+(4)^{2}}\) = 5 units.
Hence, (i) is the correct option.

Question 301.
Statement (A) : In a right angled triangle, if cos θ \(\frac{1}{2}\) = and sin θ = \(\frac{\sqrt{3}}{2}\) then tan θ = √3
‘Statement (B) : tan θ = \(\frac{\sin \theta}{\cos \theta}\)
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
Explanation:
Both A and B are correct and B is the correct explanation of the A.
tan θ = \(\frac{\sqrt{3}}{2}\) × 2 = √3
Hence, (i) is the correct option.

Question 302.
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.
Answer:
ii) A is true, B is false
Explanation:
A is correct, but B is not correct,
sin θ = cos (90 – θ)
sin 47° = cos (90 – 47)
= cos 43°
Hence, (ii) is the correct option.

❖ Study the given information and answer to the following questions.
In ΔABC, right angled at B.

AB + AC = 9 cm and BC = 3 cm

Question 303.
The value of cot C is
Answer:
\(\frac{3}{4}\)
Explanation:
cot C = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}\)
[∴ In ΔABC, By Pythagoras theorem,
AC² = AB² + BC²
AB = 4 cm, AC = 5 cm]

Question 304.
The value of sec C is
Answer:
\(\frac{5}{3}\)
Explanation:
sec C = \(\frac{A C}{B C}=\frac{5}{3}\)

Question 305.
sin² C + cos² C is equal to
Answer:
1
In figure, ΔABC has a right angle at B. If AB = BC = 1cm and AC = √2 cm.

Explanation:
sin C= \(\frac{4}{5}\) ;cosC = \(\frac{3}{5}\)
L.H.S. = sin² C + cos² C
= \(\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}\)
= \(\frac{16+9}{25}\) = 1 = R.H.S.

Question 306.
Find sin C.
Answer:
\(\frac{1}{\sqrt{2}}\)
Explanation:
sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}\)

Question 307.
Find cos C.
Answer:
\(\frac{1}{\sqrt{2}}\)
Explanation:
cos C =\(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}\)

Question 308.
Find tan C.
Answer:
1
The length of a pendulum is 80 cm. Its end describes an arc of length 16 cm.
Explanation:
tan C = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{1}\) = 1

Question 309.
To find the length of the arc which formula is useful?
Answer:
l = r0

Question 310.
Calculate the angle of arc makes at centre.
Answer:
θ = \(\frac{1}{r}=\frac{16}{80}=\frac{1}{5}\)
In ΔPQR, right angled at Q,
PR + QR = 25 cm and PQ = 5 cm

Question 311.
Determine the value of “QR”.
Answer:
QR = 12 cm

Question 312.
Determine the value of “PR”.
Answer:
PR = 13 cm

Question 313.
Find the value of sin P.
Answer:
\(\frac{12}{13}\) cm

Question 314.
Find the value of cos P.
Answer:
\(\frac{5}{13}\) cm

Question 315.
Find the value of tan P.
Answer:
\(\frac{12}{5}\) cm

Question 316.
In ΔABC, ∠B = 90°, AB = 3 cm and BC = 4 cm, then match the column.
A) sin C [ ] i) 3/5
B) tanA [ ] ii) 4/5
C) cos C [ ] iii) 5/3
D) sec A [ ] iv) 4/3
Answer:
A – (i), B – (iv), C – (ii), D – (iii)

Question 317.
Match the following.

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

Question 318.
Match the following.

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

Question 319.
If sin θ = \(\frac{7}{25}\), then

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

Question 320.
Match the following.

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

Question 321.
What is the value of
sec 16°- cosec 74° – cot 74° • tan 16° ?
Answer:
1 (one)

Question 322.
If x = 2019°, then what is the value of sin² x + cos² x ?
Solution:
If x = 2019°, then
sin²x + cos²x = sin²2019° + cos²2019° = 1 [∵ sin²θ + cos²θ = 1]

Question 323.
If x is in first quadrant and sin x = cos x, then what is the value of x?
Solution:
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°