If A, B, C are the interior angles of a triangle ABC, prove that

Solution:

(i) We have to prove: $\tan \left(\frac{C+A}{2}\right)=\cot \frac{B}{2}$

Since we know that in triangle $A B C$

$A+B+C=180$

$\Rightarrow C+A=180^{\circ}-B$

$\Rightarrow \frac{C+A}{2}=90^{\circ}-\frac{B}{2}$

$\Rightarrow \tan \frac{C+A}{2}=\tan \left(90^{\circ}-\frac{B}{2}\right)$

$\Rightarrow \tan \frac{C+A}{2}=\cot \frac{B}{2}$

Proved

 (ii) We have to prove: $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

Since we know that in triangle $A B C$

$A+B+C=180$

$\Rightarrow B+C=180^{\circ}-A$

$\Rightarrow \frac{B+C}{2}=90^{\circ}-\frac{A}{2}$

$\Rightarrow \sin \frac{B+C}{2}=\sin \left(90^{\circ}-\frac{A}{2}\right)$

$\Rightarrow \sin \frac{B+C}{2}=\cos \frac{A}{2}$

Proved