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प्रश्न
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
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उत्तर
Given cot θ + tan θ = x and sec θ – cos θ = y
x = cot θ + tan θ
x = `1/tan theta + tan theta`
= `(1 + tan^2 theta)/tan theta`
= `(sec^2 theta)/tan theta`
= `(1/cos^2theta)/(sin theta/costheta`
= `1/(cos theta sin theta)`
y = sec θ – cos θ
= `1/cos theta - cos theta`
= `(1 - cos^2 theta)/cos theta`
y = `(sin^2 theta)/costheta`
= `[1/(cos^2thetasin^2theta) xx (sin^2theta)/costheta]^(2/3) - [1/(cos theta sin theta) xx (sin^4 theta)/(cos^2 theta)]^(2/3)`
= `[1/(cos^3theta)]^(2/3) - [(sin^3 theta)/(cos^3 theta)]^(2/3)`
= `[1/(cos^2 theta)] - [(sin^2 theta)/(cos^2 theta)]`
= `[(1 - sin^2 theta)/(cos^2 theta)]`
= `[(cos^2 theta)/(cos^2 theta)]`
= 1
L.H.S = R.H.S
⇒ `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
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