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Trigonometry, sin^2alpha *sec^2beta +tan^2 beta *cos^2alpha=sin^2alpha+tan^...

sin^2alpha *sec^2beta +tan^2 beta *cos^2alpha=sin^2alpha+tan^2 beta

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Show that af+bd+ce=ae+bf+cd= 1/2 , In figure, the incircle of triangle ABC...

In figure, the incircle of triangle ABC touches the sides BC, CA, and AB at D, E, and F respectively. Show that AF+BD+CE=AE+BF+CD= 1/2   (perimeter of triangle ABC), Ans:

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