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Question: If \(\cos x=\tan y,\cot y=\tan z\) and \(\cot z=\tan x\) then \(\sin x=\) A.\(\dfrac{\sqrt{5}+1}{...

If cos⁡x=tan⁡y,cot⁡y=tan⁡z\cos x=\tan y,\cot y=\tan z and cot⁡z=tan⁡x\cot z=\tan x then sin⁡x=\sin x=
A.5+14\dfrac{\sqrt{5}+1}{4}
B. 5−14\dfrac{\sqrt{5}-1}{4}
C. 5+12\dfrac{\sqrt{5}+1}{2}
D. 5−12\dfrac{\sqrt{5}-1}{2}

Explanation

Solution

The basic formulae of trigonometry, cot⁡x=cos⁡xsin⁡x,tan⁡x=sin⁡xcos⁡x\cot x=\dfrac{\cos x}{\sin x},\tan x=\dfrac{\sin x}{\cos x}
(cot⁡x)×(tan⁡x)=1,sin⁡2x+cos⁡2x=1(\cot x)\times (\tan x)=1,{{\sin }^{2}}x+{{\cos }^{2}}x=1.

Complete step by step answer:
It is clear from the question that we have three different types of equations with trigonometric functions like cos⁡x,tan⁡y,cot⁡y,tan⁡z,cot⁡z,tan⁡x\cos x,\tan y,\cot y,\tan z,\cot z,\tan x

& \cos x=\tan y-(i) \\\ & \cot y=\tan z-(ii) \\\ & \cot z=\tan x-(iii) \\\ \end{aligned}$$ We know that $\cot x.\tan x=1$similarly $(\cot z)\times (\tan z)=1$ and $(\cot y)\times (\tan y)=1$ From the above three equations we will have $(\cos x)\times (\tan z)\times (\cot z)=(\tan y)\times (\cot y)\times (\tan x)$ Here we already know that, $(\cot z)\times (\tan z)=1$ and $(\cot y)\times (\tan y)=1$ $\cos x\times 1=1\times \tan x$ $\Rightarrow \cos x=\tan x$ Now, from the basic trigonometric formulae we have that $\tan x=\dfrac{\sin x}{\cos x}$ $\begin{aligned} & \Rightarrow \cos x=\dfrac{\sin x}{\cos x} \\\ & \Rightarrow {{\cos }^{2}}x=\sin x \\\ \end{aligned}$ As, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ Substituting the value of ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ in ${{\cos }^{2}}x=\sin x$ $\begin{aligned} & \Rightarrow 1-{{\sin }^{2}}x=\sin x \\\ & \Rightarrow {{\sin }^{2}}x+\sin x-1=0 \\\ \end{aligned}$ For a quadratic equation $a{{x}^{2}}+bx+c$ the roots will be $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ${{\sin }^{2}}x+\sin x-1=0$ this is a equation quadratic in sinx a=1, b=1, c= -1 the roots will be, substituting the values of a, b, c in $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ $\Rightarrow $ sinx= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$=$\dfrac{-1\pm \sqrt{{{(1)}^{2}}-4(1)(-1)}}{2(1)}$ $\Rightarrow $ sinx=$\dfrac{-1\pm \sqrt{1+4}}{2}=\dfrac{-1\pm \sqrt{5}}{2}$ $\Rightarrow $ sinx= $\,\dfrac{\sqrt{5}-1}{2},\dfrac{-\sqrt{5}-1}{2}$ Our required value of sinx would be sinx=$\dfrac{\sqrt{5}-1}{2}$ **So, the correct answer is “Option D”.** **Note:** Here in this question we have been provided our options with direct numerical values. But there might be cases where we will be given options with trigonometric values like $\sin 18{}^\circ ,\cos 54{}^\circ ,\sin 72{}^\circ $ which have some of the standard numerical values as there solution.