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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II3 Marks
Question
$\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$
✓

Answer

$\begin{aligned}
& \text { In } \triangle A B C \\
& A+B+C=\pi \\
& \therefore A+B=\pi-C \\
& \therefore \quad \tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{\pi-C}{2}\right) \\
& \therefore \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\cot \frac{C}{2}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \therefore \quad \tan \frac{C}{2} \cdot\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=1-\tan \frac{A}{2} \cdot \tan \frac{B}{2} \\
& \therefore \quad \tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{A}{2} \cdot \tan \frac{C}{2}+\tan \frac{A}{2} \cdot \tan \frac{B}{2}=1
\end{aligned}$
Dividing throughout by $\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2} \cdot \tan \frac{\mathrm{C}}{2}$, we get
$\begin{aligned}
& \frac{1}{\tan \frac{\mathrm{A}}{2}}+\frac{1}{\tan \frac{\mathrm{B}}{2}}+\frac{1}{\tan \frac{C}{2}}=\frac{1}{\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2} \cdot \tan \frac{\mathrm{C}}{2}} \\
& \therefore \quad \cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{B}}{2} \cdot \cot \frac{\mathrm{C}}{2}
\end{aligned}$

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