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प्रश्न
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
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उत्तर
LHS = `tan^3 θ/(1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ)`
= `tan^3 θ/sec^2 θ + cot^3 θ/(cosec^2 θ)`
= 1 + tan2θ = sec2θ; 1 + cot2θ = cosec2θ
= `sin^3 θ/cos^3 θ xx cos^2 θ + cos^3 θ/sin^3 θ xx sin^2 θ`
= `sin^3 θ/cos θ + cos^3 θ/sin θ`
= `(sin^4 θ + cos^4 θ)/(cos θ.sin θ)`
= `((sin^2θ)^2 + (cos^2θ)^2)/(sin θ.cos θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ. cos^2 θ)/(sin θ.cos θ)` ...[a2 + b2 = (a + b)2 − 2ab]
= `((1)^2 - 2sin^2θ. cos^2θ)/(sin θ.cos θ)`
= `(1 - 2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= `1/(sinθ.cosθ) - (2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= secθ. cosecθ − 2 sinθ cosθ
= RHS
Hence proved.
संबंधित प्रश्न
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
cos4 A − sin4 A is equal to ______.
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
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.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
