Question

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

Answer 1

Hint: The basic formulae of trigonometry, $\cot x=\dfrac{\cos x}{\sin x},\tan x=\dfrac{\sin x}{\cos x}$
$(\cot x)\times (\tan x)=1,{{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Complete step by step answer:
It is clear from the question that we have three different types of equations with trigonometric functions like $\cos x,\tan y,\cot y,\tan z,\cot z,\tan x$
\[\begin{align}
  & \cos x=\tan y-(i) \\
 & \cot y=\tan z-(ii) \\
 & \cot z=\tan x-(iii) \\
\end{align}\]
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 $\tan x=\dfrac{\sin x}{\cos x}$
$\begin{align}
  & \Rightarrow \cos x=\dfrac{\sin x}{\cos x} \\
 & \Rightarrow {{\cos }^{2}}x=\sin x \\
\end{align}$
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 ${{\cos }^{2}}x=\sin x$
$\begin{align}
  & \Rightarrow 1-{{\sin }^{2}}x=\sin x \\
 & \Rightarrow {{\sin }^{2}}x+\sin x-1=0 \\
\end{align}$
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.