The value of `(2(sin2theta+2cos^2theta-1))/(costheta-sintheta-cos3theta+sin3theta)` is `adot\ costheta` b. `sectheta` c. `cosectheta` d. `sintheta`

Explanation

We know that `cos A-cos B=2 sin(frac{A+B}{2})sin(frac{A-B}{2})`

`sin A-sin B=2cos(frac{A+B}{2})sin(frac{A-B}{2})`

`cos^2theta=2 cos^2 theta-1`

`(2(sin2theta+2cos^2theta-1))/(costheta-sintheta-cos3theta+sin3theta)=frac{2(sin 2theta+cos^2theta)}{2 sin(frac{theta+3 theta}{2})sin(frac{3 theta-theta}{2})+2cos(frac{3theta+theta}{2})sin(frac{3theta-theta}{2})}`

`=frac{sin 2theta+cos 2theta}{sin 2 thetasin theta+cos 2theta sin theta}`

`=frac{1}{sin theta}`

`=cosec theta`