Question

If $x+y=\frac{\pi }{4}$ and $\tan x+\tan y=1,$ then $(n\in Z)$

• A. $\sin x=0$ always
• B. when $x=n\pi +\frac{\pi }{4}$ then $y=-n\pi$
• C. when $x=n\pi$ then $y=-n\pi +\frac{\pi }{4}$
• D. when $x=n\pi +\frac{\pi }{4}$ then $y=n\pi -\frac{\pi }{4}$

Answer 1

Answer: (B), (C)

The correct options are
B when $x=n\pi +\frac{\pi }{4}$ then $y=-n\pi$
C when $x=n\pi$ then $y=-n\pi +\frac{\pi }{4}$

$x+y=\frac{\pi }{4}$
Taking $\tan$ on both sides,
$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=1=\frac{1}{1-\tan x\tan y}$ $($Given, $\tan x+\tan y=1)$
$1-\tan x\tan y=1$
$\tan x\tan y=0$
$\tan x=0$ or $\tan y=0$
$\Rightarrow \tan y=1$ or $\Rightarrow \tan x=1$
$x=n\pi ,y=-n\pi +\frac{\pi }{4}$ or $y=-n\pi ,x=n\pi +\frac{\pi }{4}$

Hence, options $B$ and $C$ are correct.