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Question
Prove the following identities:
`(tan θ + sec θ - 1)(tan θ + sec θ + 1) = (2 sin θ)/(1 - sin θ)`
Theorem
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Solution
LHS = (tan θ + sec θ)2 – 1 = tan2θ + sec2θ + 2 sec θ tan θ – 1
= tan2θ + (1 + tan2θ) + 2 sec θ tan θ – 1
= 2 tan2θ + 2 sec θ tan θ
= 2 tan θ (tan θ + sec θ)
= `2 · (sin θ)/(cos θ) · ((sin θ)/(cos θ) + 1/(cos θ))`
= `(2 sin θ(1 + sin θ))/(cos^2θ)`
= `(2 sin θ(1 + sin θ))/((1 - sin^2θ))`
= `(2 sin θ)/((1 - sin θ))`
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