If A+B=90 prove that under root tanA tanB+cotA tanB/cosA cscB-sec^2A/csc^2B=cotA
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tanA tanB + cot A tan B ./ cos A csc B =tan B( tan A + cot A) /cosA csc B
tan A + cot A = 1 / sin A cos A , tan B / cos A csc B = sin 2 B /cosA cosB
since A + B = 90, cos B = sin A Hence, LHS becomes
sin 2 B / sin 2A cos 2 A - sin 2 B/ cos 2 A
now sin 2 B = cos 2 A, I / sin 2 A - 1 = cos 2 A / sin 2 A
taking square root, LHS = cot A = RHS
tan A + cot A = 1 / sin A cos A , tan B / cos A csc B = sin 2 B /cosA cosB
since A + B = 90, cos B = sin A Hence, LHS becomes
sin 2 B / sin 2A cos 2 A - sin 2 B/ cos 2 A
now sin 2 B = cos 2 A, I / sin 2 A - 1 = cos 2 A / sin 2 A
taking square root, LHS = cot A = RHS
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