rebelmime rebelmime
  • 13-01-2021
  • Mathematics
contestada

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

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Dead3ill
Dead3ill Dead3ill
  • 13-01-2021

Answer:

[tex] \frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y[/tex]

Step-by-step explanation:

We know that [tex] { \tan }^{2} y = { \sec }^{2} y - 1[/tex]

Also , [tex] { \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y } [/tex]

So ,

[tex] \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[/tex]

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