If $A+B+C=180^{\circ}$, then the value of $\tan \left(\frac{\mathrm{A}}{2}\right) \tan…

If $A+B+C=180^{\circ}$, then the value of $\tan \left(\frac{\mathrm{A}}{2}\right) \tan \left(\frac{\mathrm{B}}{2}\right)+\tan \left(\frac{\mathrm{B}}{2}\right) \tan \left(\frac{\mathrm{C}}{2}\right)+\tan \left(\frac{\mathrm{C}}{2}\right) \tan \left(\frac{\mathrm{A}}{2}\right)$ is
  1. 1
  2. $-1$
  3. $-2$
  4. 2

Solution

$\ln \Delta A B C, A+B+C=\pi \Rightarrow A+B=\pi-C$ $\tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{\pi-C}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)$ $\therefore \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2} \Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$ $\therefore \tan \frac{\mathrm{C}}{2}\left(\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}\right) \quad=1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}$ $\therefore \tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}+\tan \frac{\mathrm{B}}{2} \tan \frac{\mathrm{C}}{2}+\tan \frac{\mathrm{C}}{2} \tan \frac{\mathrm{A}}{2}=1$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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