Question

If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1

Answer 1

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