The value of 3 cosec20^°- 20^° is (a) 2 (b) 2 20^° 40^° (c) 4 (d) 4 20^° 40^°

Correct answer: (c) 4 ( 3)/( 20^°)- 1 20^° =( 3 20^°- 20^°)/( 20^° 20^°)=(2sin(60^°-20^°))/( 12 40^°)=(2 40^°)/( 12 40^°)=4. JEE Advanced 1988 · Trigonometry — verified solution by the Vidaara Team.

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[JEE Advanced 1988] the value of sqrt 3 cosec 20 sec 20 is

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The value of $\sqrt3\,\operatorname{cosec}20^\circ-\sec20^\circ$ is

(a) $2$
(b) $\dfrac{2\sin20^\circ}{\sin40^\circ}$
(c) $4$
(d) $\dfrac{4\sin20^\circ}{\sin40^\circ}$

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (c) $4$

$\frac{\sqrt3}{\sin20^\circ}-\frac1{\cos20^\circ}=\frac{\sqrt3\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}=\frac{2\sin(60^\circ-20^\circ)}{\frac12\sin40^\circ}=\frac{2\sin40^\circ}{\frac12\sin40^\circ}=4.$

JEE Advanced 1988 · Trigonometry — verified solution by the Vidaara Team.

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