Question

If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`

Answer 1

If `sqrt(3)` sin θ – cos θ = θ

To prove tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`

`sqrt(3)` sin θ – cos θ = θ

`sqrt(3)` sin θ = cos θ 

`sin theta/cos theta = 1/sqrt(3)`

tan θ = tan 30°

θ = 30°

L.H.S = tan 3θ°

= tan3 (30°)

= tan 90°

= undefined (α)

R.H.S = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`

= `(3tan30^circ - tan^2 30^circ)/(1 - 3tan^2 30^circ)`

= `3(1/sqrt(3)) - (1/sqrt(3))^3 ÷ 1 - 3 xx (1/sqrt(3))^2`

= `sqrt(3) - 1/(3sqrt(3)) ÷ 1 - 3 xx 1/3`

= `(9 - 1)/(3sqrt(3)) ÷ 1 - 1`

= `8/(3sqrt(3)) ÷ 0`

= undefined (α)

∴ tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`