Let PQR be a triangle of area `Delta` with `a=2,b=7/2 and c=5/2`, where a,b and c are the lengths of the sides of the triangle opposite to the angles at P,Q and R respectively. Then `((2sinP-sin2P)/ (2sinP+sin2P))` equals

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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