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Question
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
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Solution
sin3θ + cos3θ
= (sin θ + cos θ)(sin2θ + cos2 – sin θ cos θ)
= (sin θ + cos θ)(1 – sin θ cos θ). ...(i)
L.H.S = `(sin^3θ + cos^3θ)/(sinθ + cosθ) + sinθcosθ`
= `((sinθ + cosθ)(1 - sinθcosθ))/((sinθ + cosθ)) + sinθcosθ` ...(From(i))
= 1 – sin θ cos θ + sin θ.cos θ
Simplify by cancelling – sin θ cos θ and + sin θ.cos θ
= 1
= R.H.S.
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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
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
