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

##### 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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