Practice the AP 10th Class Maths Bits with Answers Chapter 11 Trigonometry on a regular basis so that you can attempt exams with utmost confidence.

## AP SSC 10th Class Maths Bits 1st Lesson Chapter 11 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)

⇒ cosec^{2} θ – cot^{2} 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 tan^{2} θ – sec^{2} θ.

Answer:

-1

Explanation:

-(sec^{2} θ – tan^{2} 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 x^{2} – y^{2}.

Answer:

4

Explanation:

sec^{2} θ = \(\left(\frac{x}{2}\right)^{2}\),tan^{2} θ = \(\left(\frac{y}{2}\right)^{2}\)

⇒ sec^{2} θ – tan^{2} θ = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)

⇒ 4 = x^{2} – y^{2}

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) sin^{2} 20° + sin^{2} 70° = 1

(II) log_{2} (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:

sin^{2}A

Explanation:

\(\sqrt{\frac{\sin ^{2} A}{\operatorname{cosec}^{2} A}}\) = \(\sqrt{\sin ^{4} A}\) = sin^{2}A

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.

sin^{2}47° + sin^{2}43° 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.

sec^{4}A – sec^{2}A equal to

Answer:

tan^{4} A – tan^{2} A

Explanation:

sec^{4} A – sec^{2}A

⇒1 + tan^{4} A – (1 + tan^{2} A)

⇒ tan^{4} A – tan^{2} A

Question 48.

Find the value of \(\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}\).

Answer:

sin^{2} 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 + tan^{2} θ)cos^{2} 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.

sec^{2} θ + cosec^{2} θ equal to

Answer:

sec^{2} θ.cosec^{2} θ

Explanation:

Question 62.

sec^{2} 33° – cot^{2} 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:

2sec^{2} 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,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = 5^{2} + 12^{2}

= 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 3sin^{2} 45°+2cos^{2} 60°.

Answer:

2

Explanation:

Question 89.

Find the value of tan^{2} 30° + 2 cot^{2} 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 sec^{2} A – tan^{2} A.

Answer:

1

Explanation:

sin A = \(\frac{3}{5}\) ⇒ sec^{2}A = tan^{2}A = 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 cos^{2} 17° – sin^{2} 73°.

Answer:

0

Explanation:

cos^{2} 17°- sin^{2} 73°

= cos^{2} (90 – 73) – sin^{2} 73°

= sin^{2} 73° – sin^{2} 73° = 0

Question 96.

Find the value of sin^{2} 60° – sin^{2} 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:

– tan^{4}A

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 (k^{2} – \(\frac{1}{\mathbf{k}^{2}}\))

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 cosec^{2} A – 1.

Answer:

\(\frac{225}{64}\)

Explanation:

tan A = \(\frac{8}{15}\),

cosec ^{2} A – 1 = cot^{2}A = \(\left(\frac{15}{8}\right)^{2}=\frac{225}{64}\)

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.

cos^{2} θ equal to

Answer:

1 – sin^{2} θ

Question 111.

If tan A = \(\frac{3}{4}\) then find sec^{2} A – tan^{2} A.

Answer:

1

Question 112.

sin^{4} θ – cos^{4} θ equal to

Answer:

2sec^{2} 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 x^{2} – y^{2}.

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.

tan^{2} Φ – sec^{2} 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 cos^{2}17° – sin^{2} 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:

-sin^{2} 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 + sin^{2} θ = 1, then find cos^{2} θ + cos^{4} 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.

sin^{2} 75° + cos^{2} 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.

cot^{2}θ – \(\frac{1}{\sin ^{2} \theta}\) equal to

Answer:

-1

Explanation:

cot^{2} θ – cosec^{2} 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.

cos^{6} θ + sin^{6} θ equal to .

Answer:

1 – 3 sin^{2} θ.cos^{2} θ

Explanation:

cos^{6} θ + sin^{6} θ = (cos^{2} θ)^{3} + (sin^{2} θ)^{3}

a^{3} + b^{3} = (a + b) – 3ab (a + b)

= (sin^{2} θ + cos^{2} θ)^{3}

– 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)

= 1 – 3 sin^{2} θ cos^{2} θ (1)

= 1 – 3 sin^{2} θ cos^{2} 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 x^{2} – y^{2}.

Answer:

4

Explanation:

\(\frac{\mathrm{x}}{2}\) = cosec θ, \(\frac{\mathrm{y}}{2}\) = cot θ

cosec^{2} θ – cot^{2} θ = \(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2}\right)^{2}\)

1 = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)

x^{2} – y^{2} = 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 cot^{2} 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.

sin^{2} 45° + cos^{2} 45° + tan^{2} 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 + tan^{2} A) (1 – sin^{2} A)

equal to

Answer:

1

Explanation:

sec^{2} A × cos^{2} 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 tan^{2} θ + cot^{2} θ.

Answer:

2

Explanation:

tan θ + cot θ = 2

⇒ tan^{2} θ + cot^{2} θ + 2 . tan θ . cot θ = 4

⇒ tan^{2} θ + cot^{2} 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(a^{2} + b^{2})

= 2 (sin^{2} θ + cos^{2} θ) = 2

Question 201.

(1 + tan θ)^{2} equal to

Answer:

sec^{2} 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.

sin^{3} θ cos θ.cos^{3} 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 sin^{2} θ + 3 cos^{2} θ.

Answer:

4

Explanation:

Question 209.

cot (270° – θ) equal to

Answer:

tan θ

Question 210.

(1 + tan^{2} 60°)^{2} equals to

Answer:

16

Explanation:

[1 + (√3)^{2}]^{2} = (1 + 3)^{2} = 4^{2} = 16

Question 211.

sin (270° + θ) equal to

Answer:

– cos θ

Question 212.

If tan^{2} 60° + 2 tan^{2} 45° = x tan 45°, then find x.

Answer:

5

Explanation:

(√3)^{2} + 2(1)^{2} = x . 1

⇒ 3 + 2 = x ⇒ x = 5

Question 213.

cos^{2} 0° + cos^{2} 60° equal to

Answer:

5/4

Question 214.

Simplify : sin^{4} θ – cos^{4} θ

Answer:

2sin^{2} θ – 1

Question 215.

If α + β = 90° and α = 2β, then find cos^{2} β + sin^{2} β.

Answer:

1

Explanation:

α = 90 – β

⇒ 90 – β = 2β ⇒ 3β = 90° ⇒ β = 30°

∴ cos^{2} 30° + sin^{2} 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 = sin^{2} 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 + cos^{2} θ.

Answer:

\(\frac{5}{2}\)

Explanation:

sin θ = cos θ ⇒ θ = 45°

2 tan 45° + cos^{2} 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 – sec^{2} θ) (1 – cosec^{2} 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 sec^{2} A – tan^{2} 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.

(sec^{2} θ – 1)(1 – cosec^{2} 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 +cot^{2} 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 \(\frac{1}{2}\) tan^{2} 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 (tan^{2} θ – sec^{2} 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 4sin^{2} θ + tan^{2} 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): sin^{2} 67° + cos^{2} 67° = 1

Statement (B) : For any value of θ,

sin^{2} θ + cos^{2} 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:

sin^{2} θ + cos^{2} θ = 1

⇒ sin^{2} 67° + cos^{2} 67° = 1

Hence, (i) is the correct option.

Question 296.

Statement (A): If cos A + cos^{2} A = 1,

then sin^{2} A + sin^{4} A = 2

Statement (B): 1 – sin^{2} A = cos^{2} 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 + cos^{2} A = 1

cos A = 1 – cos^{2} A = sin^{2} A

∴ sin^{2} A + sin^{4} A = cos A + cos^{2} A = 1

⇒ sin^{2} A + sin^{4} A = 1

Hence, (iii) is the correct option.

Question 297.

Statement (A):

The value of sec^{2} 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 sec^{2} 10° – cot^{2} 80°

= sec^{2} 10°-cot^{2} (90°- 10°)

= sec^{2} 10° – tan^{2} 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

= sin^{2} θ + cos^{2} 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,

AC^{2} = AB^{2} + BC^{2}

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.

sin^{2} C + cos^{2} 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. = sin^{2} C + cos^{2} 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 sin^{2} x + cos^{2} x ?

Solution:

If x = 2019°, then

sin^{2}x + cos^{2}x = sin^{2}2019° + cos^{2}2019° = 1 [∵ sin^{2}θ + cos^{2}θ = 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°