A) 1
B) -1
C) 2
D) None of these
Correct Answer: A
Solution :
| [a] since \[A+B+C=\pi \] |
| \[\therefore \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi }{2}\Rightarrow \frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}\] |
| \[\therefore \tan \left( \frac{A}{2}+\frac{B}{2} \right)=\tan \left( \frac{\pi }{2}-\frac{C}{2} \right)=\cot \frac{C}{2}\] |
| \[\Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}\] |
| \[\Rightarrow \tan \frac{A}{2}\tan \frac{C}{2}+\tan \frac{B}{2}\tan \frac{C}{2}=1-\tan \frac{A}{2}\tan \frac{B}{2}\] |
| \[\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1\] |
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