Q72 / 125 · BITSAT 2021 · Mathematics
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$ then
- A$\tan A \tan B+\tan B \tan C+\tan C \tan A=1$
- B$\cot A+\cot B+\cot C=\cot A \cot B \cot C$
- C$\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}=1+4 \sin \mathrm{A} \sin \mathrm{B} \sin C$
- Correct answerDAll three are correct
Solution
$\mathrm{B}+\mathrm{C}=\frac{\pi}{2}-\mathrm{A} \Rightarrow \tan (\mathrm{B}+\mathrm{C})=\cot \mathrm{A}$
$\Rightarrow \frac{\tan \mathrm{B}+\tan \mathrm{C}}{1-\tan \mathrm{B} \tan \mathrm{C}}=\cot \mathrm{A}$
$\Rightarrow \tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan \mathrm{A}=1$
$\Rightarrow \tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}(\cot \mathrm{C}+\cot \mathrm{A}+\cot \mathrm{B})=1$
$\Rightarrow \cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}=\cot \mathrm{A} \cot \mathrm{B} \cot \mathrm{C}$
Again $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}$
$=2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2 \mathrm{C}$
$=2 \cos \left(\frac{\pi}{2}-C\right) \cos (A-B)+1-2 \sin ^{2} C$
$=2 \sin C[\cos (A-B)-\sin C]+1$
$=2 \sin \mathrm{C}\left[\cos (\mathrm{A}-\mathrm{B})-\sin \frac{\pi}{2}-(\mathrm{A}+\mathrm{B})\right]+1$
$\begin{array}{l}
=2 \sin C[\cos (A-B)-\cos (A+B)]+1 \\
=4 \sin A \sin B \sin C+1
\end{array}$
Hence, the correct answer is option: 'D'.