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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)`
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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`)
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
