Question

The value of $\tan (x+y)$, if none of the angles $x,y$ and $(x+y)$ is an odd multiple of $\frac{\pi }{2}$ is

• A. $\frac{\tan x-\tan y}{1-\tan x\cdot \tan y}$
• B. $\frac{\tan x+\tan y}{1-\tan x\cdot \tan y}$
• C. $\frac{\tan x+\tan y}{1+\tan x\cdot \tan y}$
• D. $\frac{1+\tan x\cdot \tan y}{\tan x+\tan y}$

Answer 1

Answer: (B)

Since, none of the $x,y$ and $(x+y)$ is an odd multiple of $\frac{\pi }{2}$, it follows that $\cos x,\cos y$ and $\cos (x+y)$ are non-zero. Now,
$\tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$
On dividing numerator and denominator by $\cos x\cos y$, we have
$\tan (x+y)=\frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}=\frac{\tan x+\tan y}{1-\tan x\tan y}$