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Q.
In a $\Delta ABC$, if $A = tan^{-1}\, 2$ and $B = tan^{ -1}\, 3$ , then $C =$
Inverse Trigonometric Functions
Solution:
We have, $A=tan^{-1}\,2 \Rightarrow tan\,A=2$
and $B=tan^{-1}\,3 \Rightarrow tan\,B=3$
since $A, B, C$ are angles of a triangle, $A+B+C = \pi$
$\Rightarrow C=\pi-\left(A+B\right)$
Now, $A+B=tan^{-1}\,2+tan^{-1}\,3=\pi+tan^{-1}\left[\frac{2+3}{1-2.3}\right]$
$\left[\therefore tan^{-1}\,x+tan^{-1}\,y=\pi+tan^{-1}\left[\frac{x+y}{1-xy}\right]
for \,x>0, y>0\,and\,xy>1\right]$
$=\pi+tan^{-1}\left(-1\right)=\pi-tan^{-1}\,1=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
$\therefore From \left(1\right), C=\pi-\frac{3\pi}{4}=\frac{\pi}{4}$