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प्रश्न
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
उत्तर
LHS = sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) (sin4θ + cos4θ - sin2θ⋅cos2θ)
= (1)[(sin2θ + cos2θ)2 - 2sin2θ⋅cos2θ - sin2θ⋅cos2θ]
= (1)[(1)2 - 3sin2θ⋅cos2θ]
= 1 - 3sin2θ ⋅ cos2θ
= RHS
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
