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Q. If $A + B + C =\frac{\pi}{2}$ then

BITSATBITSAT 2021

Solution:

$B+C=\frac{\pi}{2}-A \Rightarrow \tan (B+C)=\cot A$
$\Rightarrow \frac{\tan B+\tan C}{1-\tan B \tan C}=\cot A$
$\Rightarrow \tan A \tan B+\tan B \tan C+\tan C \tan A=1$
$\Rightarrow \tan A \tan B \tan C(\cot C+\cot A+\cot B)=1$
$\Rightarrow \cot A+\cot B+\cot C=\cot A \cot B \cot C$
Again $\cos 2 A+\cos 2 B+\cos 2 C$
$=2 \cos (A+B) \cos (A-B)+\cos 2 C$
$=2 \cos \left(\frac{\pi}{2}-C\right) \cos (A-B)+1-2 \sin ^{2} C$
$=2 \sin C[\cos (A-B)-\sin C]+1$
$=2 \sin C\left[\cos (A-B)-\sin \frac{\pi}{2}-(A+B)\right]+1$
$=2 \sin C[\cos (A-B)-\cos (A+B)]+1$
$=4 \sin A \sin B \sin C+1$