Trigonometrical Ratios And Identities Ques 15
The value of the expression $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to
$(1988,2 M)$
(a) 2
(b) $2 \sin 20^{\circ} /\sin 40^{\circ}$
(c) 4
(d) $4 \sin 20^{\circ} /\sin 40^{\circ}$
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Answer:
Correct Answer: 15.(c)
Solution:
Formula:
- Given expression $=$
$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\tan 60^{\circ} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$
$ =\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \cdot \sin 20^{\circ}}{\cos 60^{\circ} \cdot \sin 20^{\circ} \cdot \cos 20^{\circ}} $
$ =\frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\cos 60^{\circ} \cdot \sin 20^{\circ} \cdot \cos 20^{\circ}}=\frac{\sin 40^{\circ}}{\frac{1}{2} \cdot \sin 20^{\circ} \cos 20^{\circ}} $
$ =\frac{2 \sin 20^{\circ} \cos 20^{\circ}}{\frac{1}{2} \sin 20^{\circ} \cos 20^{\circ}}=4$
BETA
