NIMCET 2020 — Mathematics PYQ

Concept:

Trigonometric Ratios:

 

$\tan \theta$=$\frac{\sin \theta }{\cos \theta }$		$\cot \theta =\frac{\cos \theta }{\sin \theta }$
$\csc \theta =\frac{1}{\sin \theta }$	$\sec \theta =\frac{1}{\cos \theta }$		$\cot \theta =\frac{1}{\tan \theta }$
${\sin }^2\theta +{\cos }^2\theta =1$

The solution to the quadratic equation $a{x}^2+bx+c=0$ is given by: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:
It is given that $\cos x=\tan y$, $\cot y=\tan z$ and $\cot z=\tan x$.
∴ $\cos x=\tan y=\frac{1}{\cot y}=\frac{1}{\tan z}=\cot z=\tan x=\frac{\sin x}{\cos x}$
⇒ ${\cos }^{2}x=\sin x$
⇒ $1-{\sin }^{2}x=\sin x$
⇒ ${\sin }^{2}x+\sin x-1=0$

Using the formula for the roots of a quadratic equation:
⇒ $\sin x=\frac{-1\pm \sqrt{1^2-4(1)(-1)}}{2(1)}$
⇒ $\sin x=\frac{-1+\sqrt{5}}{2}$ OR $\sin x=\frac{-1-\sqrt{5}}{2}$

From the given answer options, $\sin x=\frac{\sqrt{5}-1}{2}$.