If `alpha` is a root of `25cos^2theta+5costheta-12=0,pi/2ltalphaltpi` the `sin2alpha` is equal to:

Options

A. `(-3)/(4)`
B. `(3)/(4)`
C. `(-4)/(3)`
D. `(-4)/(5)`

Answer

A

Explanation

Here `alpha` is the root of equation
`25cos^(2)theta+5costheta-12=0`
`rArr25cos^(2)alpha+5cosalpha-12=0`
`rArr25cos^(2)alpha+20cosalpha-15cosalpha-12=0`
`rArr5cosalpha(5cosalpha+4)-3(5cosalpha+4)=0`
`(5cosalpha-3)(5cosalpha+4)=0`
`cosalpha=(3)/(5)orcosalpha=(-4)/(5)`
Here , `(pi)/(2)ltalphaltpi`
`thereforecosalpha=(-4)/(5)`
`(because"In" 2^("nd")="quadrant,cos " alpha " value is negative")`
Now , `sinalpha=sqrt(1-cos^(2)alpha)=sqrt(1-(16)/(25))`
`thereforesinalpha=(3)/(5)`
`thereforetanalpha=(sinalpha)/(cosalpha)=(3)/(5)xx(-5)/(4)=(-3)/(4)`
`therefore` Option (a) is correct .