A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
Given that, \[A+B+C={{180}^{o}}\] \[A+B={{180}^{o}}-C\] \[\Rightarrow \frac{A}{2}+\frac{B}{2}={{90}^{o}}-\frac{C}{2}\] \[\Rightarrow \tan \left( \frac{A}{2}+\frac{B}{2} \right)=\tan \left( {{90}^{o}}-\frac{C}{2} \right)\] \[\Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\cot \frac{C}{2}\] \[\Rightarrow \left( \tan \frac{A}{2}+\tan \frac{B}{2} \right)\tan \frac{C}{2}=1-\tan \frac{A}{2}\tan \frac{B}{2}\] \[\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}\] \[+\tan \frac{A}{2}\tan \frac{C}{2}=1\]
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