NIMCET 2018 — Mathematics PYQ

Concept:
Trigonometric Identities:
- $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.
- $\tan (n\pi +\theta )=\tan \theta$.
- $\tan (-\theta )=-\tan \theta$

Calculations:

It is given that $\frac{\tan x}{2}=\frac{\tan y}{3}=\frac{\tan z}{5}=$k(say).

$\therefore \tan$x=2k,tan y=3k and tan z=5k.

It is also given that x+y+z=π.

$\Rightarrow \mathbf{x}+\mathbf{y}=\pi -\mathbf{z}$

$\Rightarrow \tan \left(\mathrm{x}+\mathrm{y}\right)=\tan \left[\pi +\left(-z\right)\right]$
$\Rightarrow \frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x}\tan \mathrm{x}}=\tan (-$z$)=-\tan$z

$\Rightarrow \tan$x$+\tan$y$=-\tan z+\tan$x$\tan$y$\tan$z$\to \tan$x$+\tan$y$\tan$

$\Rightarrow \tan$x$+\tan$y$+\tan$z$=\tan$x$\tan$y$\tan$z

Substituting the values in terms of k from the above result, we get.

$2\mathbf{k}+3\mathbf{k}+5\mathbf{k}=(2\mathbf{k})(3\mathbf{k})(5\mathbf{k})$

$\Rightarrow 10$k$=30{\mathrm{k}}^3$

$\Rightarrow {\mathrm{k}}^2=\frac{1}{3}.$

Now,  ta$n^2$x+ ta$n^2$y+ ta$n^2$z
=(2k${)}^2+(3$k${)}^2+(5$k${)}^2$
$=4{\mathrm{k}}^2+9{\mathrm{k}}^2+25{\mathrm{k}}^2$
$=38{\mathrm{k}}^2$

$=\frac{38}{3}.$