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  • Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (1 + tan^2 A /1+cot^2 A) = (1 - tan A/1- cot A)^2=tan^2 A

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A
 

Answers (1)

best_answer

We need to prove,
(\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A

Taking LHS;

\\\Rightarrow \frac{1+\tan ^{2}A}{1+\cot ^{2}A} = \frac{\sec^2A}{\csc^2A}=\tan^2A

Taking RHS;

\\=(\frac{1-\tan A}{1-\cot A})^2\\\\= (\frac{1-\sin A/\cos A }{1-\cos A /\sin A})^2\\\\ = \frac{(\cos A -\sin A)^2(\sin^2A)}{(\sin A-\cos A)^2(\cos^2A)}\\\\ =\tan^2A

LHS = RHS

Hence proved.

Posted by

manish

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