Prove that - cos-sin 90o-sin-cos 90o-1

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Trigonometric Ratios of Compound Angles

Prove that : ...

Question

Prove that :
$cos θ sin (90^{o} - θ) + sin θ cos (90^{o} - θ) = 1$

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cos θ sin (90^{o} - θ) + sin θ cos (90^{o} - θ) = 1

\frac{sin θ cos (90^{o} - θ) cos θ}{sin (90^{o} - θ)} + \frac{cos θ sin (90^{o} - θ) sin θ}{cos (90^{o} - θ)}

sin θ cos (90^{o} - θ) + cos θ sin (90^{o} - θ) + 4

\frac{cot θ}{csc θ + 1} + \frac{csc θ + 1}{cot θ} = 2 sec θ

\sqrt{\frac{sec θ - 1}{sec θ + 1}} + \sqrt{\frac{sec θ + 1}{sec θ - 1}} = 2 csc θ

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