NIMCET 2020 — Mathematics PYQ

Concept:
Trigonometric Identities:

- $\tan \theta =\frac{1}{\cot \theta }=\frac{\sin \theta }{\cos \theta }$
- $\csc \theta =\frac{1}{\sin \theta }$
- $\sec \theta =\frac{1}{\cos \theta }$
- $\cos \theta =\sin (90^{\circ }-\theta )$
- $\cot \theta =\tan (90^{\circ }-\theta )$
- ${\cos }^{-1}x=90^{\circ }-{\sin }^{-1}x$
- ${\cot }^{-1}x=90^{\circ }-{\tan }^{-1}x$
- $\sin 2\theta =2\sin \theta \cos \theta$
- $\cos 2\theta =({\cos }^2\theta -{\sin }^2\theta )=(2{\cos }^2\theta -1)=(1-2{\sin }^2\theta )$

Calculation:

$S=2{\tan }^{-1}[\csc ({\tan }^{-1}x)-\tan ({\cot }^{-1}x)]$

$\Rightarrow S=2{\tan }^{-1}[\csc ({\tan }^{-1}x)-\tan (90-{\tan }^{-1}x)]$

Let ${\tan }^{-1}x=\theta$

$\Rightarrow S=2{\tan }^{-1}[\csc \theta -\tan (90-\theta )]$

$\Rightarrow S=2{\tan }^{-1}[\csc \theta -\cot \theta ]$

$\Rightarrow S=2{\tan }^{-1}\left[\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }\right]$

$\Rightarrow S=2{\tan }^{-1}\left[\frac{1-\cos \theta }{\sin \theta }\right]$

$\Rightarrow S=2{\tan }^{-1}\left[\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right]$

$\Rightarrow S=2{\tan }^{-1}\left[\tan \frac{\theta }{2}\right]$

$\Rightarrow S=\theta ={\tan }^{-1}x$