Prove that (Sin (90° - θ))/Cos θ + (Tan (90° - θ))/Cot θ + (Cosec (90° - θ))/Sec θ = 3

Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.

Sum

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

= `cos θ/cos θ + cot θ/cot θ + sec θ/sec θ`

= 1 + 1 + 1

= 3
= RHS
Hence proved.

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