Question

If `cos^2alpha-sin^2alpha=tan^2beta," then prove that "tan^2alpha=cos^2beta-sin^2beta`.

Answer 1

`cos^2alpha-sin^2alpha=tan^2beta`
`rArr (cos^2alpha-sin^2alpha)/(cos^2alpha=sin^2alpha)=sin^2beta/cos^2beta`
`rArr(cos^2alpha-sin^2alpha+cos^2alpha+sin^2alpha)/(cos^2alpha-sin^2alpha-cos^2alpha-sin^2alpha)=(sin^2beta+cos^2beta)/(sin^2beta-cos^2beta)`
`rArr1/tan^2alpha=1/(cos^2beta-sin^2beta)`
`:. tan^2alpha=cos^2beta-sin^2beta`

Comments

Thank you for the solution it was helpful .