**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:

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°.

= sin25 cos65° + cos25 sin65

= sin25 cos (90 – 25) + cos25 . sin (90 – 25)

= sin25 . sin25 + cos25 cos25

= sin^{2}25 + cos^{2}25 = 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

(ii) (1 +tanθ + secθ) (1 + cotθ – cosec0) =

(A) 0

(B) 1

(C) 2

(D) – 1

(iii) (sec A + tan A) (1 – sinA) =

(A) sec A

(B) sin A

(C) cosec A

(D) cos 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:

(i) (B): Since, 9 sec^{2} A – 9 tan^{2} A

= 9 (sec^{2} A – tan^{2} A) = 9 (1) = 9 [∵ sec^{2} A – tan^{2} A = 1]

(ii) (C): Here,

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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