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Question
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
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Solution
LHS = `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
= `(cosec θ. cos θ. cot θ)/(cosec θ. cos θ. cot θ)`
= 1
= RHS
Hence proved.
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
