$$\dfrac{1+tan^2A}{1+cot^2A}$$ =

(A) secA

(B) -1

(C) cot2A

(D) tan2A

Asked by Abhisek | 1 year ago |  45

Solution :-

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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