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Prove : (1-cos^(2)A) *sec ^(2)B + tan^(2...

Prove : `(1-cos^(2)A) *sec ^(2)B + tan^(2)B (1- sin^(2)A) = sin^(2) A + tan^(2)B`

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`LHS = (1- cos^(2)A) * sec ^(2)B + tan^(2)B (1-sin^(2)A) = sin^(2) A + tan^(2)B`
` = sin^(2) A * sec^(2) B + tan^(2)B (1-sin^(2)A)`
` = sin^(2)A8 sec^(2)B + tan^(2)B - sin^(2)A* tan^(2)B " … " [{:(sin^(2)A+cos^(2)A=1),(therefore sin^(2)A = 1 - cos^(2)A):}]`
On rearranging the terms, we get
` LHS = sin^(2) A * sec^(2) A* tan^(2)B + tan^(2)B`
` = sin^(2) A (sec^(2)B - tan^(2) B) + tan ^(2)B`
` = sin^(2) A ( sec^(2) B - tan^(2) B) + tan^(2) B`
` = sin^(2) A (1) + tan^(2) B " ..." [{:(sin^(2)B=1+tan^(2) B),(therefore sec^(2)B - tan^(2)B=1):}]`
` = RHS`
` :. (1-cos^(2)A)* sec^(2)B+tan^(2)B (1-sin^(2)A) = sin^(2) A + tan^(2)B`
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