NIMCET 2020 — Mathematics PYQ

Concept:
The identities of trigonometry are:
- $\tan a=\frac{\sin a}{\cos a}$
- $\sin 2a=2\sin a\cos a$
- $\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$
- $\cos 2a=({\cos }^{2}a-{\sin }^{2}a)=(2{\cos }^{2}a-1)=(1-2{\sin }^{2}a)$

Calculation:

Let $\tan \left(45^{\circ }+\frac{\theta }{2}\right)=x$

$\Rightarrow x=\frac{\tan 45^{\circ }+\tan \frac{\theta }{2}}{1-\tan 45^{\circ }\tan \frac{\theta }{2}}$

$\Rightarrow x=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}(\because \tan 45^{\circ }=1)$

$\Rightarrow x=\frac{\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}+1}{1-\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}$

$\Rightarrow x=\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\times \frac{\cos \frac{\theta }{2}}{\cos \frac{\theta }{2}}$

$\Rightarrow x=\frac{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}{)}^2}{{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}}$

$\Rightarrow x=\frac{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}+2\cos \frac{\theta }{2}\sin \frac{\theta }{2}}{\cos \theta }$

$\Rightarrow x=\frac{1+\sin \theta }{\cos \theta }(\because {\sin }^{2}a+{\cos }^{2}a=1)$

$\Rightarrow x=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\sec \theta +\tan \theta$