Let `P Q R` be a triangle of area `` with `a=2,b=7/2,a n dc=5/2, w h e r ea , b ,a n dc` are the lengths of the sides of the triangle opposite to the angles at `P ,Q ,a n dR` respectively. Then `(2sinP-sin2P)/(2sinP+sin2P)e q u a l s`

Explanation

Here, `a = 2, b= 7/2, c = 5/2`

`:. s = (a+b+c)/2 = (2+7/2+5/2)/2 = 4`

Now, `(2sinP-sin2P)/(2sinP+sin2P) = (2sinP-2sinPcosP)/(2sinP+2sinPcosP)`

`=(2sinP(1-cosP))/(2sinP(1+cosP))`

`=(1-cosP)/(1+cosP)`

`=(2sin^2(P/2))/(2cos^2(P/2))`

`=tan^2(P/2)`

As, `tan(P/2) = sqrt(((s-b)(s-c))/(s(s-a)))`, so it becomes,

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