Question


A.

cos2A

B.

sinA

C.

cosA

D.

sin2 A

Correct option is D

Given: 
To find: value of 2tanA1+tan2A\frac{ 2 \tan A}{1 + tan^2 A} 
Formula Used: 
tanA=sinAcosA sin2A=2sinAcosA 1+tan2A=sec2A\tan A = \frac{ \sin A}{\cos A} \\ \ \\ \sin 2 A = 2 \sin A \cos A \\ \ \\ 1 + \tan^2 A = \sec^2 A 
Solution: 
=2tanA1+tan2A =2sinA/cosAsec2A =2sinAcosA =sin2A= \frac{2 \tan A}{1 + \tan^2 A} \\ \ \\ = \frac{2 \sin A / \cos A }{ \sec^2 A } \\ \ \\ = 2 \sin A \cos A \\ \ \\ = \sin 2A​​​​

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