I have this question on an assignment and my calculator won't show the horizontal asymptote correctly can I get some help here?

Mathematics

1 answer:

tester [92]2 years ago

5 0

What's the question? I can try and help..

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Answer:

$\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y$

Step-by-step explanation:

We know that ${ \tan }^{2} y = { \sec }^{2} y - 1$

Also , ${ \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y }$

So ,

$\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 } = \frac{ { \sin}^{2} y}{ { \tan }^{2} y} = \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } } = { \cos }^{2} y$