If `sinA-cosA=(sqrt(3)-1)/(2)`, then the value of `sinA.cosA` is

Options

A. `(1)/(sqrt(3))`
B. `(sqrt(3))/(2)`
C. `(1)/(4)`
D. `(sqrt(3))/(4)`

Answer

d

Explanation

`sinA-cosA=(sqrt(3)-1)/(2)`
Shotcut method:-
Put `theta=60^(@)`
`rArrsin60^(@)-cos60^(@)=sqrt(3)/(2)`
`rArrsqrt(3)/(2)-(1)/(2)=sqrt(3-1)/(2)`
`rArr(sqrt(3)-1)/(2)=(sqrt(3)-1)/(2)`
(Mathced)
Hence `SinA.cosA`
`sin60^(@).cos60^(@)`
`rArrsqrt(3)/(2)xx(1)/(2)=sqrt(3)/(4)`
`sinA-cosA=(sqrt(3)-1)/(2)`
Squaring both side,
`rArrsin^(2)A+cos^(2)A-3sinAcosA`
`=((sqrt(3)-1)/(2))^(2)`
`rArr1-2sinA cosA=(3+1-2sqrt(2))/(4)`
`rArr2sinA cosA=101((2-sqrt(3)))/(4)`
`rArr2sinA.cosA=(2-2+sqrt(3))/(2)`
`sinA.cosA=sqrt(3)/(4)`