\( \dfrac{1+tan^2A}{1+cot^2A}\) =

(A) secA                 

(B) -1              

(C) cot2A                

(D) tan2A

Asked by Abhisek | 1 year ago |  45

1 Answer

Solution :-

Right answer is  (D) tan2A

We know that,

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

Now, substitute this in the given problem, we get

1+\( \dfrac{tan^2A}{1}\)+cot2A

= (1+\( \dfrac{ 1}{cot^2A}.1\))+cot2A

= (cot2A+\( \dfrac{ 1}{cot^2A}\))×(\( \dfrac{1}{1+cot^2A}\))

\( \dfrac{ 1}{cot^2A}\) = tan2A

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

Answered by Pragya Singh | 1 year ago

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