Question

if tan thita is 3√3 find other trigonometric ratios​

Answer 1

Step-by-step explanation:

tanø = 3√3

• cotø = 1/tanø = 1/(3√3)

sec²ø = 1 + tan²ø = 1 + 27 = 28

• secø = √28

cosecø = 1 + cot²ø = 1 + 1/27 = 28/27

• cosecø = √(28/27)

• sinø = 1/cosecø = √27/28

• cosø = 1/secø = 1/√28

Answer 2

Answer:

If tan ∅ = 3√3

as we know that

tan Ø = Perpendicular/base

let Perpendicular = 3√3 x units

let base= x units

by Pythagoras theorem,

hypotenuse² = base² + Perpendicular²

h² = x² + (3√3x)²

h² = x² + 27x²

h² = 28x²

h = √28x²

h = 2√7x units

now, Sin∅ = perpendicular/base

=> 3√3x/2√7x

=> 3√3/2√7

Cos ∅ = b/h

=> x/2√7x

=> 1/2√7

Tan Ø is given 3√3

Cot∅ = 1/tan Ø

=> 1/3√3

sec ∅ = 1/Cos∅

=> 2√7

cosec ∅ = 1/Sin ∅

=> 2√7/3√3

Required trigonometric ratios are

• Sin ∅ = 3√3/2√7
• Cos ∅ = 1/2√7
• Tan Ø = 3√3
• Cot ∅ = 1/3√3
• Sec ∅ = 2√7
• Cosec ∅ = 2√7/3√3