Prove that (Sin θ. Cos (90 - θ) Cos θ)/Sin( 90 - θ) + (Cos θ Sin (90 - θ) Sin θ)/(Cos(90 - θ)) = 1

Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.

Sum

LHS = `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ))`

= `(sin θ. sin θ cos θ)/(cos θ) + (cos θ . cos θ sin θ)/(sin θ)`

= sin² θ + cos²θ
= 1
= RHS
Hence proved.

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