MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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 ]
= \(\vec{a}\) + (sin² 63° + cos² 63°)
[∴ sin² θ + cos² 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 = \(\sqrt{1-\frac{1}{4}}\) =\(\sqrt{\frac{3}{4}}\) = \(\frac{\sqrt{3}}{2}\)
[∴ sin² A + cos² 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 = \(\frac{a}{b}\)
[ ∴ sin² θ + cos² 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 + sin² A = 1,then the value of the expression (cos² A + cos⁴ A) is:

(A) 1
(B) \(\frac{1}{2}\)
(C) 2
(D) 3
Answer:
(A) 1

Explanation:
Given, sin A + sin² A = 1
∴ sin A = 1 – sin² A= cos²A
[∴ sin2 θ + cos2 θ = 1]
On squaring both sides ,we get
sin² A = cos⁴ A
∴ 1 – cos2 A = cos4 A
∴ cos² A + cos⁴ A = 1

Question 13.

The vale of 9 sec² A – 9 tan² A is

(A) 1
(B) 9
(C) 8
(D) 0
Answer:
(B) 9

Explanation:
9 sec² A – 9 tan² A = 9(sec² A – tan² A)
= 9 (1) [∴ sec² A – tan² 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 sin² θ and y = 2 cos²θ + 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 sin²θ
and y = 2 cos² 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; sec² θ – tan² θ = 1
sec²θ = 1 + tan² θ
sec² 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 = sec² 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 = sec² θ (1 + sin θ )(1 – sin θ )
or, k + 1 = sec² θ (1 – sin²θ )
or, k + 1 = sec²θ .cos² θ
[∴ sin 2 θ + cos² θ = 1]
or, k + 1 = sec²θ × \(\frac{1}{\sec ^{2} \theta}\)
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 cot² 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 + cos²θ – 1 is 10.

Answer:
(C) A is true but R is false

Expianation :
In case of assertion:
2 cot² A – 1 = 2(cosec² A – 1) – 1
(∴ cot ² θ = – 1 + cosec²θ
= \(\frac{2}{\sin ^{2} A}\) – 3
= \(\frac{2}{\left(\frac{\sqrt{3}}{2}\right)^{2}}\) – 3
∴ 2cot² 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 + cos² θ – 1 = sin θ -(1 – cos²θ
= sin θ – sin²θ
(∴ sin² θ + cos²∴ = 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 cot² 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 cot² 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° + cos² 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 cot²A – 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 cot² 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 ,
AB² = BC ²+ AC²
⇒ (25 – x²) = x² + (5)²
⇒ 625 – 50 x + x² + 25
⇒ 50x = 600
⇒ x = \(\frac{600}{50}\) = 12
Hence, BC = 12 cm

MCQ Questions for Class 10 Maths with Answers