Prove that (Tan θ)/(Cot(90° - θ)) + (Sec (90° - θ) Sin (90° - θ))/(Cosθ. Cosec θ) = 2.

Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.

Sum

LHS = `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ)`

= `(tan θ)/(tan θ) + ( cosec θ. cos θ)/(cosθ. cosec θ)`

= 1 + 1 = 2
= RHS
Hence proved.

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