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Question
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
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Solution
LHS = `(sec^2θ - sin^2θ)/tan^2θ`
= `(1/cos^2θ - sin^2θ)/(sin^2θ/cos^2θ)`
= `(1 - sin^2θcos^2θ)/((cos^2θ)/(sin^2θ/cos^2θ))`
= `(1 - sin^2θcos^2θ)/sin^2θ`
= `1/sin^2θ - (sin^2θcos^2θ)/(sin^2θ)`
= `cosec^2θ - cos^2θ`
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RELATED QUESTIONS
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cos θ . sec θ = ?
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Activity:
L.H.S. = `square`
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