Question

Show that tan 2A/(1+Sec2A) = tan A​

Answer 1

Answer:

tan A = sin A /cos A

tan A = 2 sin A cos A /2 cos ^2 A

tan A = sin 2 A /cos ^2 A + cos ^2 A

tan A = sin 2 A /1 − sin ^2 A + cos ^2 A

tan A = sin 2 A/ 1 + cos 2 A  

tan A =( sin 2 A ) / cos 2 A /( 1 + cos 2 A ) / cos 2 A

tan A = tan 2 A/ 1 + sec 2 A

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Step-by-step explanation:

Answer 2

Answer:

We know that,

tan2A=2tanA/1-tan^2A

cos2A=1-tan^2A/1+tan^2A

secA=1/cosA

A/Q:

{2tanA/(1-tan^2A)}/[1+1+tan^2A/1-tan^2A}]

2tanA/(1-tan^2A)}/[1+1+tan^2A/1-tan^2A}]{2tanA/(1-tan^2A)}/{1-tan^2A+1+tan^2A/(1-tan^2A)}

2tanA/2

tanA

Hence proved