Question

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

• A. $x=n\pi$ always satisfies the equation
• B. If $x=n\pi +\frac{\pi }{4},$ then $y=-n\pi$
• C. If $x=n\pi ,$ then $y=-n\pi +\frac{\pi }{4}$
• D. If $x=n\pi +\frac{\pi }{4},$ then $y=-n\pi +\frac{\pi }{4}$

Answer 1

Answer: (B), (C)

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

From $\tan x+\tan y=1$ we have $\frac{\sin x}{\cos x}+\frac{\sin x}{\cos y}=1$
$\Rightarrow \sin x\cos y+\sin y\cos x=\cos x\cos y$
$\Rightarrow \sin (x+y)=\cos x\cos y$
$\Rightarrow 2\sin (x+y)=\cos (x+y)+\cos (x-y)$
$\Rightarrow 2\sin \frac{\pi }{4}=\cos \frac{\pi }{4}+\cos (x-y)$
$\Rightarrow \cos (x-y)=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$
$\Rightarrow x-y=2n\pi \pm \frac{\pi }{4},n\in Z...\left(1\right)$
Also we have $x+y=\frac{\pi }{4}...\left(2\right)$
From Eqs. $\left(1\right)$ and $\left(2\right)$, we have
$x=n\pi +\frac{\pi }{4}\text{and}y=-n\pi ,n\in Z$
$\text{or}x=n\pi \text{and}y=-n\pi +\frac{\pi }{4},n\in Z$