## Introduction to Trigonometry Class 10 MCQ Questions with Answers

Question 1.

## The value of the expression [cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35° – 0)] is:

(A) -1

(B) 0

(C) 1

(D) \(\frac{3}{2}\)

Answer:

(B) 0

Explanation:

cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35°-0)

= cosec [90° – (15° – 0)] – sec(15° – 0)

– tan(55° + 0) + cot[90° – (55° + 0)]

= sec(15° – 0) – sec(15° – 0) – tan(55° + 0)

+ tan(55° + 0)

= 0

Question 2.

## If cos (α + ß) = 0, then sin (α – ß) can be reduced to

(A) cos ß

(B) cos 2ß

(C) sin α

(D) sin 2α

Answer:

(B) cos 2ß

Explanation:

cos (α + ß) = 0

cos(α + ß) = cos 90°

α + ß = 90°

a = 90° – ß

sin (α – ß) = sin (90° – ß – ß

= sin (90° – 2ß)

= cos 2ß

Question 3.

## The value of (tan 1° tan 2° tan 3°… tan 89°) is:

(A) 0

(B) 1

(C) 2

(D) \(\frac{1}{2}\)

Answer:

(B) 1

Explanation:

(tan 1° tan 2° tan 3° … tan 89°)

= (tan 1° tan 89°)(tan 2° tan 88°)(tan 3° tan 87°)…(tan 45° tan 45°)

= [tan 1° tan (90°-l)][tan 2° tan (90° -2)][tan 3°tan (90°-3)] … [tan 45° tan (90°-45°)] = tan l°cot 1° tan 2°cot 2° tan 3°cot 3°…tan 45°cot 45°

= tan 1° × \(\frac{1}{\tan 1^{\circ}}\)tan 2°. \(\frac{1}{\tan 2^{\circ}}\)tan 3°. \(\frac{1}{\tan 3^{\circ}}\) … \(\frac{\tan 45^{\circ}}{\tan 45^{\circ}}\)

= 1 . 1 . 1 . 1…… 1 . 1

= 1

Question 4.

## If cos 9α = sin a and 9α < 90°, then the value of tan 5α is:

(A) \(\frac{1}{\sqrt{3}}\)

(B) √ 3

(C) 1

(D) 0

Answer:

(C) 1

Explanation:

cos 9α = sin α

cos 9α = cos (90° – α )

On comparing both sides, we have

9α = 90° – α

10 α = 90°

α = 9°

∴ tan 5 × 9° = tan 45° = 1

Question 5.

## If AABC is right angled at C, then the value of cos (A+B) is:

(A) 0

(B) 1

(C) \(\frac{1}{2}\)

(D) \(\frac{\sqrt{3}}{2}\)

Answer:

(A) 0

Explanation:

We know that, in ∆ABC

sum of three angles = 180°

i.e., ∠ A + ∠B + ∠C = 180°

∠C = 90°

∠A +∠ B + 90° = 180°

A + B = 90°

∴ cos(A + B) = cos 90° = 0

Question 6.

## Given that sin θ = \(\frac{1}{2}\) and cos ß = \(\frac{1}{2}\), then the value of (∴ + ß) is:

(A) 0°

(B) 30°

(C) 60°

(D) 90°

Answer:

(D) 90°

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

[ ∴ sin 30° = \(\frac{1}{2}\)]

∴ θ = 30°

And, cosß = \(\frac{1}{2}\) = cos 60°

∴ ß = 60°

∴ θ + ß = 30° + 60° = 90°

[ ∴ cos 60° = \(\frac{1}{2}\)

Question 7.

## The value of the expression \(\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]\) is:

(A) 3

(B) 2

(C) 1

(D) 0

Answer:

(B) 2

Explanation:

\(\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]\)

\(=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2}\left(90^{\circ}-68^{\circ}\right)+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}\) + cos 63° sin (90° – 63°)

= \(\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\cos ^{2} 68^{\circ}+\sin ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ}\)

[∴ sin(90° – θ ) = cos θ and cos (90° – θ ) = sin θ ]

= \(\vec{a}\) + (sin^{2} 63° + cos^{2} 63°)

[∴ sin^{2} θ + cos^{2} θ = 1]

= 1 + 1 = 2

Question 8.

## If 4 tan θ = 3, then \(\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)\) is equal to:

(A) \(\frac{2}{3}\)

(B) \(\frac{1}{2}\)

(C) \(\frac{1}{3}\)

(D) \(\frac{3}{4}\)

Answer:

(B) \(\frac{1}{2}\)

Explanation:

Given, 4 tan θ = 3

∴ tan θ = \(\frac{3}{4}\)

∴ \(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\) = \(4 \frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}\)

[Divided by cos θ in both numerator and denominator]

= \(\frac{4 \tan \theta-1}{4 \tan \theta+1}\)[ ∴ [tanθ = \(\frac{\sin \theta}{\cos \theta}\)]

= \(\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}\) = \(\frac{3-1}{3+1}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) [ Put tan ∴ = \(\frac{3}{4}\) frome equation (i)

Question 9.

## If cos A =\(\frac{4}{5}\) then the value of tan A is:

(A) \(\frac{3}{5}\)

(B) \(\frac{3}{4}\)

(C) \(\frac{4}{3}\)

(D) \(\frac{1}{8}\)

Answer:

(B) \(\frac{3}{4}\)

Explanation:

Given,

cosA = \(\frac{4}{5}\)

∴ sin A = \(\sqrt{1-\cos ^{2} A}\)

[[∴ sin2 A + cos2 A = 1 ∴ sin A = \(\left.\sqrt{1-\cos ^{2} A}\right]\)

sin A = \(\sqrt{1-\left(\frac{4}{5}\right)^{2}}\) = \(\sqrt{1-\frac{16}{25}}\) = \(\sqrt{\frac{9}{25}}\) = \(\frac{3}{5}\)

tan A = \(\frac{\sin A}{\cos A}\)

= \(\frac{\frac{3}{5}}{\frac{4}{5}}\) =\(\frac{3}{4}\)

Question 10.

## If sin A = \(\frac{1}{2}\) then the value of cot A is:

(A) √3

(B) \(\frac{1}{\sqrt{3}}\)

(C) \(\frac{\sqrt{3}}{2}\)

(D) 1

Answer:

(A) √ 3

Explanation:

Given, sin A = \(\frac{1}{2}\)

cos A = \(\sqrt{1-\sin ^{2} A}\) = \(\sqrt{1-\left(\frac{1}{2}\right)^{2}}\)

cos A = \(\sqrt{1-\frac{1}{4}}\) =\(\sqrt{\frac{3}{4}}\) = \(\frac{\sqrt{3}}{2}\)

[∴ sin^{2} A + cos^{2} A = 1 θ cos A = \(\sqrt{1-\sin ^{2} A}\)]

Now, cot A = \(\frac{\cos A}{\sin A}\) = \(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\) = θ 3

Question 11.

## Given that sin θ = \(\frac{a}{b}\) then cos θ is equal to

(A) \(\frac{b}{\sqrt{b^{2}-a^{2}}}\)

(B) \(\frac{b}{a}\)

(C) \(\frac{\sqrt{b^{2}-a^{2}}}{b}\)

(D) \(\frac{a}{\sqrt{b^{2}-a^{2}}}\)

Answer:

(C) \(\frac{\sqrt{b^{2}-a^{2}}}{b}\)

Explanation:

Given, sin θ = \(\frac{a}{b}\)

[ ∴ sin^{2} θ + cos^{2} θ = 1 θ cos θ = \([\left.\sqrt{1-\sin ^{2} \theta}\right]\)

cos θ = \(\sqrt{1-\left(\frac{a}{b}\right)^{2}}\) = \(\sqrt{1-\frac{a^{2}}{b^{2}}}\) = \(\frac{\sqrt{b^{2}-a^{2}}}{b}\)

Question 12.

## If sin A + sin^{2} A = 1,then the value of the expression (cos^{2} A + cos^{4} A) is:

(A) 1

(B) \(\frac{1}{2}\)

(C) 2

(D) 3

Answer:

(A) 1

Explanation:

Given, sin A + sin^{2} A = 1

∴ sin A = 1 – sin^{2} A= cos^{2}A

[∴ sin2 θ + cos2 θ = 1]

On squaring both sides ,we get

sin^{2} A = cos^{4} A

∴ 1 – cos2 A = cos4 A

∴ cos^{2} A + cos^{4} A = 1

Question 13.

## The vale of 9 sec^{2} A – 9 tan^{2} A is

(A) 1

(B) 9

(C) 8

(D) 0

Answer:

(B) 9

Explanation:

9 sec^{2} A – 9 tan^{2} A = 9(sec^{2} A – tan^{2} A)

= 9 (1) [∴ sec^{2} A – tan^{2} A = 1]

= 9

Question 14

## The value of (1 + tan θ + sec θ )( 1 + cot θ – cosec θ ) is

(A) 0

(B) 1

(C) 2

(D) -1

Answer:

Explanation:

(1 + tan θ + sec θ )( 1 + cot θ – cosecθ )

= \(\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\)\(\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)\)

= \(\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\)\(\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right)\)

= \(\frac{(\sin \theta+\cos \theta)^{2}-(1)^{2}}{\sin \theta \cos \theta}\)

= \(\frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}\)

= \(\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}\)

= \(\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}\)

= 2

Question 15.

## The value of (sec A + tan A)(1 – sin A) is

(A) sec A

(B) sin A

(C) cosec A

(D) cos A

Answer:

(D) cos A

Explanation:

(sec A + tan A) (1 – sin A)

= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\) (1 – sin A)

= \(\left(\frac{1+\sin A}{\cos A}\right)\) (1 – sin A)

= \(\left(\frac{1-\sin ^{2} A}{\cos A}\right)\) = \(\frac{\cos ^{2} A}{\cos A}\)

= cos A

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false

(D) A is false and R is True

Question 1.

## Assertion (A): Cot A is the product of Cot and A.

Reason (R): The value of sin0 increases as 0 increases.

Answer:

(D) A is false and R is True

Explanation:

In case of assertion:

cot A is not the product of cot and A. It is the cotangent of θ A.

∴ Assertion is incorrect.

In case of reason:

The value of sin 0 increases as 0 increases in interval of θ °< θ ° < 90° ∴ Reason is correct: Hence, assertion is incorrect but reason is correct. Question 2. Assertion (A): The value of \(\frac{\tan 47^{\circ}}{\cot 43^{\circ}}\) is 1. Reason (R): The value of the expression (sin 80° – cos 80°) is negative. Answer: (C) A is true but R is false Explanation: In case of assertion: = \(\frac{\tan 47^{\circ}}{\cot 43^{\circ}}\) = \(\frac{\tan 47^{\circ}}{\cot \left(90^{\circ}-47^{\circ}\right)}\) = \(\frac{\tan 47^{\circ}}{\tan 47^{\circ}}\) = 1 ∴ Assertion is correct. In case of reason: 80° is near to 90°, sin 90° = 1 and cos 90° = 0 So, the given expression sin 80° – cos 80° > 0 So, the value of the given expression is positive.

∴ Reason is incorrect:

Hence, assertion is correct but reason is incorrect.

Question 3.

## Assertion (A): If tan A = cot B, then the value of (A + B) is 90°.

Reason (R): If sec θ sin θ = 0, then the value of θ is 0°.

Answer:

(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:

In case of assertion:

tan A = cot B (Given)

tan A = tan(90° – B)

[∴ tan (90° – θ ) = cot θ ] A = 90° – B

A + B =90°.

∴ Assertion is correct.

In case of reason:

Given, sec θ .sin θ = 0

\(\vec{a}\) = 0

or, tan θ = 0 = tan 0°

∴ θ = 0°

∴ Reason is correct:

Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 4.

## Assertion (A): If x = 2 sin^{2} θ and y = 2 cos^{2}θ + 1 then the value of x + y = 3.

Reason (R): If tan θ = \(\frac{5}{12}\) , then the value of sec θ is \(\frac{13}{12}\)

Answer:

(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:

In case of assertion:

we have x = 2 sin^{2}θ

and y = 2 cos^{2} θ + 1

Then, x + y = 2 sin2 θ + 2cos2 + 1

= 2(sin2 θ + cos2 θ ) + 1

= 2 × 1 + 1[∴ sin2 θ + cos2 θ = 1]

= 2 + 1 = 3

∴ Assertion is correct.

In case of reason:

tan θ = \(\frac{5}{12}\)

Unsing identity; sec^{2} θ – tan^{2} θ = 1

sec^{2}θ = 1 + tan^{2} θ

sec^{2} θ = 1 + \(\left(\frac{5}{12}\right)^{2}\)

= 1 + \(\frac{25}{144}\)

= \(\frac{144+25}{144}\)

= \(\frac{169}{144}\)

= \(\sqrt{\frac{13^{2}}{12^{2}}}\)

= \(\frac{13}{12}\)

∴ Reason is correct: Hence, both assertion but reason is not the assertion.

Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 5.

## Assertion (A): If k + 1 = sec^{2} θ (1 + sin θ ) (1 – sin θ ), then the value of k’ is 1.

Reason (R): If sin θ + cos θ =θ 3 then the value of tan θ + cot θ is 1.

Answer:

(D) A is false and R is True

Explanation:

In case of assertion:

k + 1 = sec^{2} θ (1 + sin θ )(1 – sin θ )

or, k + 1 = sec^{2} θ (1 – sin^{2}θ )

or, k + 1 = sec^{2}θ .cos^{2} θ

[∴ sin 2 θ + cos^{2} θ = 1]

or, k + 1 = sec^{2}θ × \(\frac{1}{\sec ^{2} \theta}\)

or, k + 1 = 1

or, k = 1 – 1

∴ k = 0.

∴ Assertion is correct.

In case of reason:

Given sin θ + cosθ = θ 3

sin2 θ + cos2 θ + 2 sin θ cos θ = 3

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 2

sin θ cos θ = 1 …..(i)

tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}\) + \(\frac{\cos \theta}{\sin \theta}\)

= \(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}\)

= \(\frac{1}{\cos \theta \sin \theta}\)

= \(\frac{1}{1}\) = 1 [From equation (i)]

∴ Reason is correct

Hence, Assertion is incorrect but reason is correct.

Question 6.

## Assertion (A): If sin A = \(\frac{\sqrt{3}}{2}\), then the value of 2 cot^{2} A – 1 is \(\frac{-1}{3}\)

Reason (R) : If θ be an acute angle and 5 cosec θ = 7, then the value of sin θ + cos^{2}θ – 1 is 10.

Answer:

(C) A is true but R is false

Expianation :

In case of assertion:

2 cot^{2} A – 1 = 2(cosec^{2} A – 1) – 1

(∴ cot ^{2} θ = – 1 + cosec^{2}θ

= \(\frac{2}{\sin ^{2} A}\) – 3

= \(\frac{2}{\left(\frac{\sqrt{3}}{2}\right)^{2}}\) – 3

∴ 2cot^{2} A – 1 =\(\frac{8}{3}\) – 3 = \(\frac{-1}{3}\)

∴ Assertion is correct.

In case of reason:

Given, 5 cosec θ = 7

or, cosec θ = \(\frac{7}{5}\)

sin ∴ = \(\frac{5}{7}\)

[∴ cosec θ = \(\frac{1}{\sin \theta}\)]

sin θ + cos^{2} θ – 1 = sin θ -(1 – cos^{2}θ

= sin θ – sin^{2}θ

(∴ sin^{2} θ + cos^{2}∴ = 1)

= \(\frac{5}{7}\) – \(\left(\frac{5}{7}\right)^{2}\)

= \(\frac{35-25}{49}\) = \(\frac{10}{49}\)

∴ Reason is correct

Hence, Assertion is incorrect but reason is correct.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following question on the basis of the same:

‘Skvsails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The skv sails technology

allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively

Question 1.

## In the given figure, if tan θ = cot (30° + θ ), where 0 and 30° + θ are acute angles, then the value of 0 is:

(A) 45°

(B) 30°

(C) 60°

(D) None of these

Answer:

(B) 30°

Expianation :

Given,

tan θ = cot(30° + θ )

= tan[90° – (30° + θ )]

= tan(90° – 30° – θ )

∴ tan θ = tan(60° -θ )

∴ θ = 60° – θ

∴ 2θ = 60°

∴ θ = 30°

Question 2.

## The value of tan 30°. cot 60° is:

(A) √3

(B) \(\frac{1}{\sqrt{3}}\)

(C) 1

(D) \(\frac{1}{3}\)

Answer:

(D) \(\frac{1}{3}\)

Explanation:

tan 30° x cot 60° = \(\frac{1}{\sqrt{3}}\) × \(\frac{1}{\sqrt{3}}\)

= \(\frac{1}{3}\)

Question 3.

## What should be the length of the rope of the kite sail in order to pull the ship at the angle 0 and be at a vertical height of 200 m

(A) 400 m

(B) 300 m

(C) 100 m

(D) 200 m

Answer:

(A) 400 m

Explanation:

In ∆ABC, we have 0 = 30°, AB = 200 m

Then, sin 30° = \(\frac{\text { Perpendicular}}{\text { Hypotenuse}}\)

= \(\frac{A B}{A C}\)

∴ \(\frac{1}{2}\) = \(\frac{200}{A C}\)

∴ AC = 400 m.

Question 4.

## If cos A = \(\frac{1}{2}\), then the value of 9 cot^{2} A – 1 is:

(A) 1

(B) 3

(C) 2

(D) 4

Answer:

(C) 2

Explanation:

Given, cos A = \(\frac{1}{2}\)

∴ cos A = cos 60°

∴ A = 60°

then, 9 cot^{2} A – 1 = 9(cot 60°)^{2} – 1

= 9 \(\left(\frac{1}{\sqrt{3}}\right)^{2}\) – 1

= 9 × \(\frac{1}{3}\) – 1 = 3 – 1

= 2

Question 5.

## In the given figure, the value of (sin C + cos A) is:

(A) 1

(B) 2

(C) 3

(D) 4

Answer:

(A) 1

Explanation:

We have,

AB = 200 m and AC = 400 m Proved in Question 3]

Then, sin C + cos A = \(\frac{A B}{A C}\) + \(\frac{A B}{A C}\)

= 2 × \(\frac{A B}{A C}\)

= 2 × \(\frac{200}{400}\) = 1

II. Read the following text and answer the following question on the basis of the same:

Authority wants to construct a slide in a city park for children. The slide was to be constructed for children below the age of 12 years. Authority prefers the top of the slide at a height of 4 m above the ground and inclined at an angle of 30° to the ground.

Question 1.

## The distance of AB is:

(A) 8 m

(B) 6 m

(C) 5 m

(D) 10 m

Answer:

(A) 8 m

Explanation:

∠B = 30° and

AC = 4 m

Then, sin 30° = \(\frac{A C}{A R}\)

⇒ \(\frac{1}{2}\) = \(\frac{4}{A B}\)

⇒ AB = 8 m.

Question 2.

## In value of sin ^{2} 30° + cos ^{2} 60° is:

(A) \(\frac{1}{4}\)

(B) \(\frac{1}{2}\)

(C) \(\frac{3}{4}\)

(D) \(\frac{3}{2}\)

Answer:

(B) \(\frac{1}{2}\)

sin ^{2} 30° + cos^{2} 60°= \(\left(\frac{1}{2}\right)^{2}\) + \(\left(\frac{1}{2}\right)^{2}\)

= \(\frac{1}{4}\) = \(\\frac{1}{4}vec{a}\).

= \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 3

## If cos A = \(\vec{a}\), then the value of cot^{2}A – 2 is:

(A) 5

(B) 4

(C) 3

(D) 2

Ans:

(D) 2

Explanation:

since, cos A = \(\vec{a}\)

⇒ cos A = cos 60°

⇒ A = 60°

Then 12 cot^{2} A -2 = 12(cot 60°) – 2

= 12 \(\left(\frac{1}{\sqrt{3}}\right)^{2}\) – 2

= 12 × \(\frac{1}{3}\) – 2

= 4 – 2 = 2.

Question 4

## In the given figure, the value of (sin C × cos A ) is:

(A) \(\frac{1}{3}\)

(B) \(\frac{1}{2}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{1}{5}\)

Answer:

(B) \(\frac{1}{2}\)

Explanation:

since, AC ⊥BC,

then ∠C = 90°

sin C cos A = sin 90 × \(\frac{A C}{A B}\)

= 1 × \(\frac{4}{8}\)

= \(\frac{1}{2}\)

Question 5.

## In the given figure, if AB + BC = 258 cm and AC = 5 cm, then the value of BC is:

(A) 25 cm

(B) 15 cm

(C) 10 cm

(D) 12 cm

Answer:

(D) 12 cm

Explanation:

we have, ∠C = 90

AB = BC = 25 cm and AC = 5 cm

let BC be x cm, then AB = (25 – x ) cmBy using Pythagoras theorem ,

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

⇒ (25 – x^{2}) = x^{2} + (5)^{2}

⇒ 625 – 50 x + x^{2} + 25

⇒ 50x = 600

⇒ x = \(\frac{600}{50}\) = 12

Hence, BC = 12 cm