Question

Simplify sin^2y/sec^2 y−1 to a single trigonometric function

Answer 1

`\frac{ { \sin}^(2) y}{ { \sec}^(2)y - 1 } = { \cos }^(2) y`

Explanation:

We know that
`{ \tan }^(2) y = { \sec }^(2) y - 1`

Also ,
`{ \tan}^(2) y = \frac{ { \sin }^(2) y}{ { \cos }^(2)y }`

So ,

`\frac{ { \sin }^(2)y }{ { \sec }^(2)y - 1 } = \frac{ { \sin}^(2) y}{ { \tan }^(2) y} = \frac{ { \sin }^(2)y }{ \frac{ { \sin}^(2) y}{ { \cos}^(2)y } } = { \cos }^(2) y`