Sin( 90 - theta ) = cos theta prove this using coordinate axes

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\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1

sin

θ sin (90^{\circ} - θ) - cos θ cos (90^{\circ} - θ) = 0.

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

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

\frac{cos (90 - θ)}{sin θ} + \frac{sin θ}{cos (90 - θ)} = 2

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