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Question
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
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Solution
LHS = (sin θ + cos θ)(cosec θ – sec θ)
= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`
= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`
= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`
= `(1 - 2sin^2θ)/(sinθ*cosθ)`
= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`
= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`
= cosec θ · sec θ – 2 tan θ
= RHS
Hence proved.
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