Question

if √3 sinA -- cosA =0 ,then show that

tan3A= 3 tanA - tan(cube) A /(divide by) 1--3tan ( square)A​

Answer 1

Step-by-step explanation:

√3 sinΦ - cosΦ = 0

To prove : tan 3Φ = (3tanΦ - tan³Φ)/(1 - 3tan²Φ)

Now,

√3 sinΦ - cosΦ = 0

=> √3 sinΦ = cosΦ

Dividing cosΦ both sides :

=> √3 tanΦ = 1

=> tanΦ = 1/√3

=> tanΦ = tan30°

=> Φ = 30°

Here,

tan 3Φ = (3tanΦ - tan³Φ)/(1 - 3tan²Φ)

tan 3 × 30° = {3 × 1/3√3 - (1/√3)³} /{1 - 3 × (1/√3)²}

tan 90° = {√3 - 1/3√3}/{1 - 3 × 1/3}

∞ = {√3 - 1/3√3}/{1 - 1}

∞ = {√3 - 1/3√3}/0

∞ = ∞

Hence, proved