Let PQR be a triangle of `!` area 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 P , Q and R respectively. Then , `(2sinP-sin2P)/(2sinP+sinP)` equals

Options

A. `3/(4!)`
B. `45/(4!)`
C. `(3/(4!))^(2)`
D. `(45/(4!))^(2)`

Explanation

We have,
`a=2,b=7/2` and `c=5/2`

`therefore2s=a+b+c`
`rArr2s=2+7/2+5/2rArr2s=8rArrs=4`
`(2sinP-sin2P)/(2sinP+sin2P)`
`=(2sinP-2sinPcosP)/(2sinP+2sinPcosP)`
`=(1-cosP)/(1+cosP)`
`=tan^(2)P/2=((s-b)(s-c))/(s(s-a))`
`=((s-b)(s-c)^(2))/(Delta^(2))={((s-b)(s-c))/(Delta)}^(2)={((2s-2b)(2s-2r))/(4Delta)}^(2)`
`={((8-7)(8-5))/(4Delta))^(2)=(3/(4Delta))^(2)`