A) \[\cot A\]
B) \[\cot \frac{A}{2}\]
C) \[\tan \,\,A\]
D) \[\tan \frac{A}{2}\]
Correct Answer: B
Solution :
In\[\Delta ABC\], we have \[A+B+C={{180}^{o}}\] \[\Rightarrow \] \[B+C={{180}^{o}}-A\] \[\Rightarrow \] \[\frac{B+C}{2}={{90}^{o}}-\frac{A}{2}\] \[\Rightarrow \] \[\tan \left( \frac{B+C}{2} \right)=\tan \left( {{90}^{o}}-\frac{A}{2} \right)\] \[\Rightarrow \] \[\tan \left( \frac{B+C}{2} \right)=\cot \frac{A}{2}\]
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