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If $\cos x=\tan y, \cot y=\tan z$ and $\cot z=\tan x$, then $\sin x$ equals to
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Verified Answer
The correct answer is:
$\frac{\sqrt{5}-1}{2}$
Given, $\cos x=\tan y, \cot y=\tan z$
and $\cot z=\tan x$
$\therefore \quad \cos x=\tan y$
$\Rightarrow \quad \cos x=\frac{1}{\tan z}$
$\Rightarrow \quad \cos x=\cot z$
$\Rightarrow \quad \cos x=\tan x$
$\Rightarrow \quad \cos x=\frac{\sin x}{\cos x}$
$\Rightarrow \cos ^2 x=\sin x$
$\Rightarrow 1-\sin ^2 x=\sin x$
$\Rightarrow \sin ^2 x+\sin x-1=0$
$\begin{aligned} \therefore \quad \sin x & =\frac{-1 \pm \sqrt{1-4 \times(-1)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{5}}{2} \\ \therefore \sin x=\frac{\sqrt{5}-1}{2} & \left(\because \frac{-1-\sqrt{5}}{2} < -1\right)\end{aligned}$
and $\cot z=\tan x$
$\therefore \quad \cos x=\tan y$
$\Rightarrow \quad \cos x=\frac{1}{\tan z}$
$\Rightarrow \quad \cos x=\cot z$
$\Rightarrow \quad \cos x=\tan x$
$\Rightarrow \quad \cos x=\frac{\sin x}{\cos x}$
$\Rightarrow \cos ^2 x=\sin x$
$\Rightarrow 1-\sin ^2 x=\sin x$
$\Rightarrow \sin ^2 x+\sin x-1=0$
$\begin{aligned} \therefore \quad \sin x & =\frac{-1 \pm \sqrt{1-4 \times(-1)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{5}}{2} \\ \therefore \sin x=\frac{\sqrt{5}-1}{2} & \left(\because \frac{-1-\sqrt{5}}{2} < -1\right)\end{aligned}$
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