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

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( 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 )

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