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If $A+B+C=180^{\circ}$, then $\Sigma \tan \frac{A}{2} \tan \frac{B}{2}$ is equal to
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Verified Answer
The correct answer is:
$1$
We have,
$A+B+C=\pi$
$\Rightarrow \quad A+B=\pi-C$
$\Rightarrow \quad \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$
$\Rightarrow \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)$
$\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2}$
$\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$
$\Rightarrow \quad \tan \frac{C}{2} \tan \frac{A}{2}+\tan \frac{B}{2} \tan \frac{C}{2}$
$=1-\tan \frac{A}{2} \tan \frac{B}{2}$
$\Rightarrow \quad \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
$\Rightarrow \quad \Sigma \tan \frac{A}{2} \tan \frac{B}{2}=1$
$A+B+C=\pi$
$\Rightarrow \quad A+B=\pi-C$
$\Rightarrow \quad \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$
$\Rightarrow \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)$
$\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2}$
$\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$
$\Rightarrow \quad \tan \frac{C}{2} \tan \frac{A}{2}+\tan \frac{B}{2} \tan \frac{C}{2}$
$=1-\tan \frac{A}{2} \tan \frac{B}{2}$
$\Rightarrow \quad \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
$\Rightarrow \quad \Sigma \tan \frac{A}{2} \tan \frac{B}{2}=1$
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