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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`
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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.
RELATED QUESTIONS
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
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Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
`(cosecA - 1)/(cosecA + 1) = (cosA/(1 + sinA))^2`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
What is the value of 9cot2 θ − 9cosec2 θ?
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
