## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

Choose the correct answer from the given four options (1 to 12) :

Question 1.

\({ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \) is equal to

(a) 1

(b) -1

(c) sin^{2} θ

(d) sec^{2} θ

Solution:

Question 2.

(sec^{2} θ – 1) (1 – cosec^{2} θ) is equal to

(a) – 1

(b) 1

(c) 0

(d) 2

Solution:

Question 3.

\(\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \) is equal to

(a) 2 sin^{2} θ

(b) 2 cos^{2} θ

(c) sin^{2} θ

(d) cos^{2} θ

Solution:

Question 4.

(cos θ + sin θ)^{2} + (cos θ – sin θ)^{2} is equal to

(a) -2

(b) 0

(c) 1

(d) 2

Solution:

Question 5.

(sec A + tan A) (1 – sin A) is equal to

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Solution:

Question 6.

\(\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A } \) is equal to

(a) sec^{2} A

(b) -1

(c) cot^{2} A

(d) tan^{2} A

Solution:

Question 7.

If sec θ – tan θ = k, then the value of sec θ + tan θ is

(a) \(1-\frac { 1 }{ k } \)

(b) 1 – k

(c) 1 + k

(d) \(\\ \frac { 1 }{ k } \)

Solution:

Question 8.

Which of the following is true for all values of θ (0° < θ < 90°):

(a) cos^{2} θ – sin^{2} θ = 1

(b) cosec^{2} θ – sec^{2} θ = 1

(c) sec^{2} θ – tan^{2} θ = 1

(d) cot^{2} θ – tan^{2} θ = 1

Solution:

Question 9.

If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is

(a) 0

(b) 2 sin θ cos θ

(c) 1

(d) 2 sin^{2} θ

Solution:

Question 10.

The value of cos 65° sin 25° + sin 65° cos 25° is

(a) 0

(b) 1

(b) 2

(d) 4

Solution:

Question 11.

The value of 3 tan^{2} 26° – 3 cosec^{2} 64° is

(a) 0

(b) 3

(c) -3

(d) -1

Solution:

Question 12.

The value of \(\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1 \) is

(a) -cot θ

(b) -sin^{2} θ

(c) -cos^{2} θ

(d) -cosec^{2} θ

Solution: