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प्रश्न
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
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उत्तर
sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ)(sin4θ + cos2θ - sin2θcos2θ)
= 1 x [(sin2θ)2 + (cos2θ)2 + 2sin2θ.cos2θ - 3sin2θ.cos2θ]
= (sin2θ)2 + (cos2θ)2 - 3sin2θ.cos2θ
= 1 - 3sin2θ cos2θ.
= RHS
संबंधित प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
