Question

Prove `cot^2theta/(sintheta+costheta)=(cos^2thetasintheta-cos^3theta)/(2sin^4theta-sin^2theta)`?

Answer 1

See below:

Explanation:

To help follow what's going on I'll put in red text things that have changed from the line before.

`cot^2theta/(sintheta+costheta)=(cos^2thetasintheta-cos^3theta)/(2sin^4theta-sin^2theta)`

I'll first use `cot^2theta=cos^2theta/sin^2theta` on the left side and factor the numerator and denominator on the right side. We can also see a `sintheta-costheta` term setting up in the right numerator and so I'll multiply through on the left using that term:

`((color(red)(cos^2theta/sin^2theta))/(sintheta+costheta))color(red)(((sintheta-costheta)/(sintheta-costheta)))=color(red)((cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1))`

`cos^2theta/((sin^2theta)(sintheta+costheta))((sintheta-costheta)/(sintheta-costheta))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

`(cos^2thetacolor(red)((sintheta-costheta)))/((color(red)(sin^2theta))(sintheta+costheta)color(red)((sintheta-costheta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

`(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((sin^2theta-cos^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

We can now use `sin^2theta+cos^2theta=1 => cos^2theta=1-sin^2theta`

`(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2theta-color(red)((1-sin^2theta))))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

`(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2thetacolor(red)(-1+sin^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

`(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((2sin^2theta-1)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)`

Answer 2

`LHS=(cos^2thetasintheta-cos^3theta)/(2sin^4theta-sin^2theta)`

`=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1))`
`=(cot^2theta(sintheta-costheta))/(sin^2theta-1+sin^2theta)`

`=(cot^2theta(sintheta-costheta))/(sin^2theta-(1-sin^2theta))`

`=(cot^2theta(sintheta-costheta))/(sin^2theta-cos^2theta)`

`=(cot^2theta(sintheta-costheta))/((sintheta-costheta)(sintheta+costheta)`

`=cot^2theta/(sintheta+costheta)=LHS`

Proved