Question

Question #ca510

Answer 1

See the answer below...

Explanation:

`cos2theta`
`=cos^2theta-sin^2theta`
`=(1-sin^2theta)-sin^2theta" "`[As `color(red)(sin^2theta+cos^2theta=1`]
`=1-2sin^2theta`
`=1-2/csc^2theta`
`=1-2/(1+cot^2theta)" "`[As `color(red)(csc^2theta-cot^2theta=1`]

Hence, `costheta=color(red)(|ul(bar(1-2/(1+cot^2theta)|`

Hope it helps...
Thank you...

Answer 2

The answer is `=-(1+cot^2theta)/(1-cot^2theta)`

Explanation:

`cottheta=costheta/sintheta`

`sin^2theta+cos^2theta=1`

Dividing by `sin^2theta`

`1+cot^2theta=csc^2theta`

`cos2theta=1-2sin^2theta`

`sin^2theta=(1-cos2theta)/2`

`csc^2theta=2/(1-cos2theta)`

Therefore,

`1+cot^2theta=2/(1-cos2theta)`

`1-cos2theta=2/(1+cot^2theta)`

`cos2theta=1-2/(1-cot^2theta)=(1-cot^2theta-2)/(1-cot^2theta)`

`=-(1+cot^2theta)/(1-cot^2theta)`