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Q.

If $\tan x = \cos x$ , then $\sin x =$

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a

$\frac{\sqrt{5} - 1}{2}$

b

$\frac{\sqrt{5} + 1}{2}$

c

$\frac{\sqrt{5} - 1}{4}$

d

$\frac{\sqrt{5} + 1}{4}$

answer is A.

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Detailed Solution

$\tan x = \cos x$

$\frac{\sin x}{\cos x} = \cos x$

$\sin x = \cos^{2} x$

$\sin x = 1 - \sin^{2} x$

$\sin^{2} x + \sin x - 1 = 0$

$\sin x = \frac{- 1 \pm \sqrt{1 + 4}}{2}$

$= \frac{- 1 + \sqrt{5}}{2}, \frac{- 1 - \sqrt{5}}{2}$

$\sin x = \frac{- 1 + \sqrt{5}}{2}$

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