Question

If$x+y+z=\pi ,tanxtany=2,tanx+tany+tanz=6$ then

• A. $x=m\pi +\pi /4$
• B. $y=n\pi +ta{n}^{-1}2$
• C. $z=l\pi +ta{n}^{-1}3$
• D. all are correct $(l,m,n\epsilon I)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $x=m\pi +\pi /4$
B $z=l\pi +ta{n}^{-1}3$
C $y=n\pi +ta{n}^{-1}2$
D all are correct $(l,m,n\epsilon I)$

Given $\tan x\tan y=2$ .....(1)

and $\tan x+\tan y+\tan z=6$ ....(2)

Also given $x+y+z=\pi$

$\Rightarrow x+y=\pi -z$

$\Rightarrow \tan (x+y)=-\tan z$

$\frac{\tan x+\tan y}{1-\tan x\tan y}=-\tan z$

$\Rightarrow \frac{6-\tan z}{1-2}=-\tan z$ (using (1) and (2))

$\Rightarrow \tan z=3$

$\Rightarrow \tan x+\tan y=3$ (using (2))

$\Rightarrow \tan x=1,\tan y=2$ (using (1))

So, the general solution of $\tan x=1$ is

$x=m\pi +\frac{\pi }{4}m\in I$

General solution of $\tan y=2$

$y=n\pi +{\tan }^{-1}2,n\in I$

General solution of $\tan z=3$

$\Rightarrow z=l\pi +{\tan }^{-1}3,l\in I$