Question

Question #b5658

Answer 1

See Below.

Explanation:

Identities Required:
`sin(2x)=2sin(x)cos(x)`

`sin^2(x)+cos^2(x)=1`
`sin^2(x)=1-cos^2(x)`

`cos(2x)=cos^2(x)-sin^2(x) =cos^2(x)-(1-cos^2(x))=2cos^2(x)-1`

Now solving:
`sin(2x)/sin(x)-cos(2x)/cos(x)=sec(x)`
`(2sin(x)cos(x))/sin(x)-(2cos^2(x)-1)/cos(x)=RHS`
`2cos(x)-2cos(x)+1/cos(x)=RHS`
`sec(x)=RHS`
QED