**KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4.

## Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4

Question 1.

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

cosec^{2} A – cot^{2} A = 1

cosec^{2} A = 1 + cot^{2} A

cosec^{2} A = cot^{2} A + 1

Question 2.

Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

i) sin^{2} A + cos^{2} A = 1

sin^{2} A = 1 – cos^{2} A

Question 3.

Evaluate :

i) \(\frac{\sin ^{2} 63^{\circ}+\sin 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

ii) sin 25° cos 65° + cos 25° sin 65°

Solution:

ii) sin 25° cos 65° + cos 25° sin 65°

= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)

= sin 25° sin 25° + cos 25° + cos 25°

= sin^{2} 25° + cos^{2} 25°

= 1. [∵ cos^{2} θ + sin^{2} θ = 1]

Question 4.

Choose the correct option. Justify your choice.

i) 9 sec^{2} A – 9 tan^{2} A.

A) 1

B) 9

C) 8

D) 0

Solution:

B) 9

9 sec^{2} A – 9 tan^{2} A

= 9(sec^{2} A – tan^{2} A)

= 9 × 1

= 9

ii) (1+tan θ + sec θ) (1+ cot θ- cosec θ) =

A) 0

B) 1

C) 2

D) -1

Solution:

C) 2

(1+tan θ + sec θ) (1+ cot θ- cosec θ)

iii) (sec A + tan A) (1 – sin A) =

A) sec A

B) sin A

C) cosec A

D) cos A

Solution:

D) cos A

(sec A + tan A) (1 – sin A)

iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =

A) sec^{2} A

B) -1

C) cot^{2} A

D) tan^{2} A

Solution:

D) tan^{2} A

Question 5.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

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