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Q.
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
Trigonometric Functions
Solution:
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
$\tan (x+y)=\frac{\frac{\sin x \cos y}{\cos x \cos y}+\frac{\cos x \sin y}{\cos x \cos y}}{\frac{\cos x \cos y}{\cos x \cos y}-\frac{\sin x \sin y}{\cos x \cos y}}=\frac{\tan x+\tan y}{1-\tan x \tan y}$