Question

"find the value of sin^2(90-theta)/cos^2theta"

Answer 1

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find the value of

$\frac{ \sin {}^{2} (90 - \alpha ) }{ \cos {}^{2} ( \alpha ) }$

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we know that

sin(90- x) = cosx

then ,

$\frac{ \sin {}^{2} (90 - \alpha ) }{ \cos {}^{2} ( \alpha ) }$

$= \frac{ \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) }$

$= 1$

Answer 2

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we know that

sin(90-x) = cosx

$\frac{ \sin {}^{2} (90 - \beta ) }{ \cos {}^{2} ( \beta ) }$

$= \frac{ \cos {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) }$

$= 1$

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