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Prove that `(i) (sin^(2)A cos^(2)B - cos^(2)A sin^(2) B )=(sin^(2)A- sin^(2)B) ` `(ii) (tan^(2)A sec^(2)B - sec^(2)A tan^(2)B)=(tan^(2)A- tan^(2)B) ` |
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Answer» We have `(i) LHS = (sin^(2)A cos^(2)B - cos^(2)A sin^(2) B )` `= sin^(2)A(1- sin^(2)B)-(1- sin^(2)A)sin^(2)B ` ` = sin^(2)A - sin^(2)B = RHS. ` `(ii) LHS = (tan^(2)A sec^(2)B - sec^(2)A tan^(2)B) ` ` =tan^(2)A(1+tan^(2)B) - (1+ tan^(2)A)tan^(2)B ` ` = (tan^(2)A - tan^(2)B) = RHS. ` |
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