Question

( operatorname{cosec}^{2}left(frac{B+C}{2}right)-tan ^{2} frac{A}{2}=1 )
( A+B+C=pi )
( B+C=pi-A )
( operatorname{cosec}^{2}left[frac{pi}{2}-frac{A}{2}right]-tan ^{2} frac{A}{2}=1 )
( sec ^{2} frac{A}{2}-tan ^{2} frac{A}{2}=1 )
( 1=1 )

# If A, B, C are interior angles of A ABC show that : coseca - tan2

Solution