NIMCET 2018 Question Paper with solutions

Concept:
Trigonometric Identities:
- $\tan \left(\mathit{A}+\mathit{B}\right)=\frac{\tan \mathit{A}+\tan \mathit{B}}{1-\tan \mathit{A}\tan \mathit{B}}$ .
- $\tan \left(\mathit{n}\mathit{\pi }+\mathit{\theta }\right)=\tan \mathit{\theta }$ .
- $\tan \left(-\mathit{\theta }\right)=-\tan \mathit{\theta }$
Calculations:
It is given that $\frac{\tan \mathit{x}}{2}=\frac{\tan \mathit{y}}{3}=\frac{\tan \mathit{z}}{5}=\mathit{k}$ (say).
$\therefore \tan \mathit{x}=2\mathit{k},\tan \mathit{y}=3\mathit{k}$ and $\tan \mathit{z}=5\mathit{k}$ .
It is also given that $\mathit{x}+\mathit{y}+\mathit{z}=\mathit{\pi }$ .
$\Rightarrow \mathit{x}+\mathit{y}=\mathit{\pi }-\mathit{z}$
$\Rightarrow \tan \left(\mathit{x}+\mathit{y}\right)=\tan \left[\mathit{\pi }+\left(-\mathit{z}\right)\right]$
$\Rightarrow \frac{\tan \mathit{x}+\tan \mathit{y}}{1-\tan \mathit{x}\tan \mathit{x}}=\tan \left(-\mathit{z}\right)=-\tan \mathit{z}$
$\Rightarrow \tan \mathit{x}+\tan \mathit{y}=-\tan \mathit{z}+\tan \mathit{x}\tan \mathit{y}\tan \mathit{z}$
$\Rightarrow \tan \mathit{x}+\tan \mathit{y}+\tan \mathit{z}=\tan \mathit{x}\tan \mathit{y}\tan \mathit{z}$
Substituting the values in terms of $\mathit{k}$ from the above result, we get:
$2\mathit{k}+3\mathit{k}+5\mathit{k}=\left(2\mathit{k}\right)\left(3\mathit{k}\right)\left(5\mathit{k}\right)$
$\Rightarrow 10\mathit{k}=30{\mathit{k}}^3$
$\Rightarrow {\mathit{k}}^2=\frac{1}{3}$ Now, ${\tan }^2\mathit{x}+{\tan }^2\mathit{y}+{\tan }^2\mathit{z}$
$={\left(2\mathit{k}\right)}^2+{\left(3\mathit{k}\right)}^2+{\left(5\mathit{k}\right)}^2$
$=4{\mathit{k}}^2+9{\mathit{k}}^2+25{\mathit{k}}^2$
$=38{\mathit{k}}^2$
$=\frac{38}{3}.$