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Andrews [41]
2 years ago
13

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

Mathematics
1 answer:
liubo4ka [24]2 years ago
6 0

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

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