NIMCET 2020 — Mathematics PYQ

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

Calculation:

Given 3 sin x+4 cos x=5

$\Rightarrow 3(2\sin \frac{x}{2}\cos \frac{x}{2})+4(1-2{\sin }^2\frac{x}{2})=$

$⟹6\sin \frac{\mathbf{x}}{2}\cos \frac{\mathbf{x}}{2}+$

$⟹6\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}-8{\sin }^2\frac{\mathrm{x}}{2}=1$

$\Rightarrow 6\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}-8{\sin }^{}$

Dividing the equation by${\cos }^{2}x2$

$\Rightarrow 6^{\begin{array}{c}\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}\\ {\cos }^2\frac{\mathrm{x}}{2}\end{array}-8\begin{array}{c}{\sin }^2\frac{\mathrm{x}}{2}\\ {\cos }^2\frac{\mathrm{x}}{2}\end{array}=\begin{array}{c}{\sin }^2\frac{\mathrm{x}}{2}+{\cos }^2\\ {\cos }^2\frac{\mathrm{x}}{2}\end{array}}$

$\Rightarrow 6\tan \frac{\mathbf{x}}{2}-8{\tan }^2\frac{\mathbf{x}}{2}={\tan }^2\frac{\mathbf{x}}{2}+1$

$\Rightarrow 6\tan \frac{\mathrm{x}}{2}-9{\tan }^2\frac{\mathrm{x}}{2}$