If `sinalpha=1/sqrt5 and sinbeta=3/5` , then `beta-alpha` lies in

Options

A. `[0,pi//4]`
B. `[pi//2,3pi//4]`
C. `[3pi//4,pi]`
D. none of these

Answer

A

Explanation

We have `sin alpha = 1//sqrt(5)rArr cos alpha = 2//sqrt(5)`
and `sin beta = 3//5 rArr cos beta = 4//5`
`sin (beta - alpha) = sin beta cos alpha - sin alpha cos beta`
`=(3)/(5).(2)/(sqrt(5))-(1)/(sqrt(5)).(4)/(5)=(2)/(5sqrt(5))=0.1789`
Now `sin.(pi)/(4)=(1)/(sqrt(2))=0.7071 = sin.(3pi)/(4)`
Since `0 lt 0.1789 lt 0.7071`
`therefore sin 0 lt sin (beta - alpha) lt sin.(pi)/(4) rArr 0 lt (beta - alpha) lt (pi)/(4)`
Also, `sin pi lt sin(beta-alpha)lt sin.(3pi)/(4)`
`therfore (beta-alpha)in[0,pi//4]` or `[3pi//4,pi]`