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Question
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
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Solution
Given x sin2 θ + y cos2 θ = sin θ cos θ
x sin θ = y cos θ ...(1)
x sin3 θ + y cos3 θ = sin θ cos θ
x sin θ (sin2 θ) + y cos θ (cos2 θ) = sin θ cos θ
x sin θ (sin2 θ) + x sin θ (cos2 θ) = sin θ cos θ
x sin θ (sin2 θ + cos2 θ) = sin θ cos θ
x sin θ = sin θ cos θ
x = cos θ
substitute x = cos θ in (1)
cos θ sin θ = y cos θ y = sin θ
L.H.S = x2 + y2 = cos2 θ + sin2 θ = 1
L.H.S = R.H.S
Hence it is proved.
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RELATED QUESTIONS
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`cos^2 A + 1/(1 + cot^2 A) = 1`
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tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
If x = a tan θ and y = b sec θ then
Choose the correct alternative:
1 + cot2θ = ?
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
