Advertisements
Advertisements
प्रश्न
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
Advertisements
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 that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
The value of sin2 29° + sin2 61° is
If cos \[9\theta\] = sin \[\theta\] and \[9\theta\] < 900 , then the value of tan \[6 \theta\] is
Which is not correct formula?
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
