Advertisements
Advertisements
Question
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
Advertisements
Solution
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)`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If cosA + cos2A = 1, then sin2A + sin4A = 1.
