Math, asked by Anonymous, 1 year ago

Show that tan 2A/(1+Sec2A)=tan A​

Answers

Answered by Anonymous
7

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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Answered by Anonymous
0

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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