If cos x = tan y,cot y = tan z and cot z = ta | vedclass

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(D) Given,$\cos x = 	an y$,$\cot y = 	an z$ and $\cot z = 	an x$. From the given equations,we have $	an y = \cos x$ and $\cot y = \frac{1}{	an y}

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

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(D) Given,$\cos x = 	an y$,$\cot y = 	an z$ and $\cot z = 	an x$. From the given equations,we have $	an y = \cos x$ and $\cot y = \frac{1}{	an y} = \frac{1}{\cos x}$. Since $\cot y = 	an z$,we have $	an z = \frac{1}{\cos x}$. Then $\cot z = \frac{1}{	an z} = \cos x$. Given $\cot z = 	an x$,we get $	an x = \cos x$. $\Rightarrow \frac{\sin x}{\cos x} = \cos x$ $\Rightarrow \sin x = \cos^2 x$ $\Rightarrow \sin x = 1 - \sin^2 x$ $\Rightarrow \sin^2 x + \sin x - 1 = 0$. Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $\sin x$ must be positive in the range where $	an x$ and $\cos x$ are defined as equal,we take the positive root: $\sin x = \frac{\sqrt{5}-1}{2}$.

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

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