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Question
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Solution
LHS = (sinθ + 1 + cosθ)(sinθ − 1 + cosθ). secθcosecθ
= [sin2θ − sinθ + sinθcosθ + sinθ − 1 + cosθ + sinθcosθ − cosθ + cos2θ] `1/cosθ1/sinθ ` ...(∵ secθ = `1/cosθ and cosecθ = 1/sinθ`)
= [1 + 2sinθcosθ − 1]`1/cosθ 1/sinθ`
= [2sinθcosθ]`1/cosθ1/sinθ`
= 2 = RHS
Hence proved.
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