Question

If $A,B,C$ are interior angles of $\Delta ABC$, show that :
$cose{c}^2\left(\frac{B+C}{2}\right)-{\tan }^2\frac{A}{2}=1$

Answer 1

Given :- $A+B+C=\pi$

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

$\left(\frac{A}{2}\right)=[\frac{\pi }{2}-\left(\frac{B+C}{2}\right)]$

$\Rightarrow \cos \left[\frac{A}{2}\right]=\cos [\frac{\pi }{2}-\left(\frac{B+C}{2}\right)]$

$\Rightarrow \cos \left(\frac{A}{2}\right)=\sin \left(\frac{B+C}{2}\right)$

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

$\frac{1}{{\cos }^2\left(A/2\right)}=\frac{1}{{\sin }^2\left(\frac{B+C}{2}\right)}$

${\sec }^2\left(A/2\right)=cose{c}^2\left(\frac{B+C}{2}\right)$ $\to$ identity

$1+{\tan }^2\left(\frac{A}{2}\right)=cose{c}^2\left(\frac{B+2}{2}\right)$

$cose{c}^2\left(\frac{B+C}{2}\right)-{\tan }^2\left(\frac{A}{2}\right)=1$