If cos x = tan y, cot y = tan z and cot z = tan x, then sin x is equal to

A
$\frac{\sqrt{5} + 1}{4}$
B
$\frac{\sqrt{5} - 1}{4}$
C
$\frac{\sqrt{5} + 1}{2}$
D
$\frac{\sqrt{5} - 1}{2}$

Solution:

Given ,cos x = tan y, cot y = tan z and cot z = tan x

$\begin{array}{l} ∴ \cos x = \tan y \Rightarrow \cos x = \frac{1}{\tan z} \\ \Rightarrow \cos x = \cot z \Rightarrow \cos x = \tan x \\ \Rightarrow \cos x = \sin x / \cos x \Rightarrow \cos^{2} x = \sin x \\ \Rightarrow 1 - \sin^{2} x = \sin x \Rightarrow \sin^{2} x + \sin x - 1 = 0 \\ ∴ \sin x = \frac{- 1 \pm \sqrt{1 - 4 \times (- 1)}}{2 \times 1} = \frac{- 1 \pm \sqrt{5}}{2} \\ \Rightarrow \sin x = \frac{\sqrt{5} - 1}{2} [∵ \frac{- 1 - \sqrt{5}}{2} < - 1] \end{array}$

Solution:

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