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Question
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Solution
LHS = `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ)`
= `((cos θ + sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ + sin θ) + ((cos θ - sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ - sin θ)`
= 1 - sin θ cos θ + 1 + sin θ cos θ
= 2
= RHS
Hence proved.
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RELATED QUESTIONS
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If tan θ = `13/12`, then cot θ = ?
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
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`square/square` = cosec2θ ......[Taking root on the both side]
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and sin θ = `1/("cosec" θ)`
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
