Show that tan 2A/(1+Sec2A)=tan A
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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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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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