Question

$A,B,C$ are interior angles of $△ABC$. then that is $\csc \left(\frac{A+B}{2}\right)=\sec \frac{C}{2}$ true/ false ?

• A. True
• B. False

Answer 1

The correct option is A True

Given

In $\Delta ABC$

$A+B+C=\pi$

$⟹A+B=\pi -C$

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

Applying Cosecant on both sides

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

$\csc \left(\frac{A+B}{2}\right)=\sec \left(\frac{C}{2}\right)$