Advertisements
Advertisements
प्रश्न
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
Advertisements
उत्तर
sin6A + cos6A = (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= 1 – 3 cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ......[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
sin4A + cos4A = (sin2A)2 + (cos2A)2
= (1 – cos2A)2 + (cos2A)2
= 1 – 2 cos2A + (cos2A)2 + (cos2A)2 ......[∵ (a – b)2 = a2 – 2ab + b2]
= 1 – 2 cos2A + 2 cos4A
= 1 – 2 cos2A(1 – cos2A)
= 1 – 2 cos2A sin2A
L.H.S = 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1
= 2(1 – 3 cos2A sin2A) – 3(1 – 2 cos2A sin2A) + 1
= 2 – 6 cos2A sin2A – 3 + 6 cos2A sin2A + 1
= 0
= R.H.S
∴ 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
APPEARS IN
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 m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Evaluate sin25° cos65° + cos25° sin65°
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`cosec theta sqrt(1 - cos^2 theta) = 1`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
