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$. Since $\cot y = 	an z$,we ha

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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$. Since $\cot y = 	an z$,we have $\frac{1}{	an y} = 	an z$,which implies $	an z = \frac{1}{\cos x}$. Also,$\cot z = 	an x$,so $\frac{1}{	an z} = 	an x$. Substituting $	an z = \frac{1}{\cos x}$ into $\cot z = 	an x$,we get $\cos x = 	an x$. $\cos x = \frac{\sin x}{\cos x} \implies \cos^2 x = \sin x$. Using $\cos^2 x = 1 - \sin^2 x$,we get $1 - \sin^2 x = \sin x$,or $\sin^2 x + \sin x - 1 = 0$. Solving this quadratic equation for $\sin x$ using the quadratic formula: $\sin x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$. Since $-1 \le \sin x \le 1$,we discard the negative root $\frac{-1-\sqrt{5}}{2} Thus,$\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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