Question

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

• 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

The correct option is D $\frac{\sqrt{5}-1}{2}$

cos x = tan y ...(1)

cot y = tan z ...(2)

cot z = tan x ...(3)

$\Rightarrow$ cos x=tan y

cot y = sec x

from equation 2

cot y = tan z = sec x

$\Rightarrow$ tan z=sec x

But, cot z = tan z from equ 3

so, cot x = sec x

$\Rightarrow$ tan x = cos x

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

$\sqrt{1-co{s}^{2}x}=co{s}^{2}x$

let $co{s}^{2}x=t$

$\Rightarrow \sqrt{1-t}=t$

$\Rightarrow 1-t=t^2$

$\Rightarrow t^2+t-1=0$

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

$\Rightarrow t=\frac{-1\pm \sqrt{5}}{2}$

$t=\frac{-1-\sqrt{5}}{2}$ or $t=\frac{-1+\sqrt{5}}{2}$ (possible)

So, $t=\frac{-1+\sqrt{5}}{2}$ option d is correct