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प्रश्न
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
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उत्तर
L.H.S. = `((1 + tan^2A)cotA)/(cosec^2A)`
= `(sec^2A cotA)/(cosec^2A` ...(∵ sec2 A = 1 + tan2 A)
= `(1/(cos^2A) xx (cosA)/(sinA))/(1/(sin^2A))`
= `(1/(cosA sinA))/(1/(sin^2A))`
= `sinA/cosA`
= tan A = R.H.S.
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Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
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.
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
