[JEE Advanced 1988] The value of $\sqrt3\,\operatorname{cosec}20^\circ-\sec20^\circ$ is
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
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· 1mo ago
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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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