Question

Question #39a5a

Answer 1

To prove `tan^2x-sin^2x=tan^2xsin^2x`

`tan^2x-sin^2x`
`=(sin^2x)/(cos^2x)-sin^2x` [Put tangent in terms of sine and cosine]
`=(sin^2x-sin^2xcos^2x)/cos^2x` [Common denominator]
`=(sin^2x(1-cos^2x))/cos^2x` [Factorise out the `sin^2x`]
`=(sin^2x*sin^2x)/(cos^2x)` [Since `sin^2x+cos^2x=1`, `1-cos^2x=sin^2x`]
`=tan^2xsin^2x` [Since `sin^2x/(cos^2)=tan^2x`]