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Question Number 73160 by Aditya789 last updated on 07/Nov/19

$$1+tanAtan\frac{A}{2}=tanAcot\frac{A}{2}-1=secA$$

Answered by $@ty@m123 last updated on 07/Nov/19

$$1+\tan A\tan \frac{A}{2}$$$$=1+\frac{2\tan \frac{A}{2}}{1-\tan {}^2\frac{A}{2}}.\tan \frac{A}{2}$$$$=1+\frac{{2\tan }^2\frac{A}{2}}{1-\tan {}^2\frac{A}{2}}$$$$=\frac{1-{\tan }^2\frac{A}{2}+{2\tan }^2\frac{A}{2}}{1-\tan {}^2\frac{A}{2}}$$$$=\frac{1+{\tan }^2\frac{A}{2}}{1-\tan {}^2\frac{A}{2}}$$$$=\frac{1}{\frac{1-{\tan }^2\frac{A}{2}}{1+\tan {}^2\frac{A}{2}}}$$$$=\frac{1}{\cos A}$$$$=\sec A$$$$SecondPart:Tryyourselfinsimilar$$$$manner.$$

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