Question

$A,B,C$ are interior angles of $△ABC$ then $\csc \left(\frac{A+B}{2}\right)=\sec \frac{C}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

We know that $A+B+C=\pi$

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

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

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

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

Hence proved.