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