Question

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`

Answer 1

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,
`=((s-b)(s-c))/(s(s-a))`
`=((s-b)(s-c))^2/(s(s-a)(s-b)(s-c))`
`=(1/2*3/2)^2/Delta^2`
`=(3/(4Delta))^2`
So, option `(c)` is the correct option.