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प्रश्न
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
उत्तर
LHS = `tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ)`
= `(sin^2θ/cos^2θ)/(sin^2θ/cos^2θ - 1) + (1/sin^2θ)/(1/(cos^2θ) - 1/sin^2θ)`
= `sin^2θ/(sin^2θ - cos^2θ) + (1/sin^2θ)/((sin^2θ - cos^2θ)/(cos^2θ sin^2θ))`
= `sin^2θ/(sin^2θ - cos^2θ) + cos^2θ/(sin^2θ - cos^2θ)`
= `(sin^2θ + cos^2θ)/(sin^2θ - cos^2θ) = 1/(sin^2θ - cos^2θ)` (∵`sin^2θ + cos^2θ = 1`)
Notes
θ
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