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प्रश्न
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
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उत्तर
LHS = `sin^8θ - cos^8θ`
= `(sin^4θ)^2 - (cos^4θ)^2`
= `(sin^4θ - cos^4θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)(sin^2θ + cos^2θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)((sin^2θ)^2 + (cos^2θ)^2 + 2sin^2θcos^2θ - 2sin^2θcos^2θ)`
= `(sin^2θ - cos^2θ)((sin^2θ + cos^2θ)^2 - 2sin^2θcos^2θ)`
= `(sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
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Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Prove that sin4A – cos4A = 1 – 2 cos2A.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
