Question

How do you simplity `6cot2theta-costheta` to trigonometric functions of `theta`?

Answer 1

`3(cottheta-tantheta)-costheta.`

Explanation:

`cot2theta=cos(2theta)/sin(2theta)=(cos^2theta-sin^2theta)/(2sinthetacostheta)=1/2{cos^2theta/(costhetasintheta)-sin^2theta/(sinthetacostheta)}=1/2(cottheta-tantheta)`

Hence, the Given Exp. `=3(cottheta-tantheta)-costheta.`