Question

If $\cos x=\tan y,\cot y=\tan z$ and $\cot z=\tan x$ then $\sin x$ equals to

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

Answer 1

Answer: (D)

Given, $\cos x=\tan y,\cot y=\tan z$
and $\cot z=\tan x$
$\therefore \cos x=\tan y$
$\Rightarrow \cos x=\frac{1}{\tan z}$
$\Rightarrow \cos x=\cot z$
$\Rightarrow \cos x=\tan x$
$\Rightarrow \cos x=\frac{\sin x}{\cos x}$
$\Rightarrow {\cos }^{2}x=\sin x$
$\Rightarrow 1-{\sin }^{2}x=\sin x$
$\Rightarrow {\sin }^{2}x+\sin x-1=0$
$\therefore \sin x=\frac{-1\pm \sqrt{1-4\times (-1)}}{2\times 1}$
$=\frac{-1\pm \sqrt{5}}{2}$
$\therefore \sin x=\frac{\sqrt{5}-1}{2}\left(\because \frac{-1-\sqrt{5}}{2}<-1\right)$