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

Asked by Abhisek | 1 year ago |  138

1 Answer

Solution :-

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

L.H.S. = \( \dfrac{ 1+tan^2A}{1+cot^2A}\)

Since cot function is the inverse of tan function,

\( \dfrac{1+tan^2A}{1}+ \dfrac{1}{tan^2A}\)

\( \dfrac{1+tan^2A}{1+tan^2A}\)

Now cancel the 1+tan2A terms, we get

= tan2A

(1+\( \dfrac{ tan^2A}{1+cot^2A}\)) = tan2A

Similarly,

\( ( \dfrac{1-tan A}{1-cot A})^2\) = tan2A

Hence proved

Answered by Pragya Singh | 1 year ago

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