Question

Prove: TanA+SecA = root3

Answer 1

Step-by-step explanation:

Hello,

it's a trig equation

secA + tanA = √3

let's rewrite secA and tanA in terms of sinA, cosA:

(1 /cosA) + (sinA /cosA) = √3

let's move √3 to the left side:

(1 /cosA) + (sinA /cosA) - √3 = 0

let cosA be the common denominator at the left side:

[1 + sinA - (√3)cosA] /cosA = 0

let's equate the numerator to zero:

1 + sinA - (√3)cosA = 0

let's group terms as:

1 + [sinA - (√3)cosA] = 0

let's multiply and divide the expression in brackets by 2:

1 + 2 {[(sinA)/2] - [(√3)cosA]/2} } = 0

1 + 2 {sinA (1/2) - [(√3)/2] cosA} = 0

given that 1/2 = cos(60°) and (√3)/2 = sin(60°), we can rewrite the equation as:

1 + 2 [sinA cos(60°) - sin(60°) cosA] = 0

let's apply, to both products in brackets, the product-to-sum formula

sin α cos β = [sin(α + β) + sin(α - β)] /2:

1 + 2 { {[sin(A + 60°) + sin(A - 60°)] /2} - {[sin(60° + A) + sin(60° - A)] /2} } = 0

1 + 2 { {sin(A + 60°) + sin(A - 60°) - [sin(A + 60°) + sin(60° - A)]} /2} = 0

1 + 2 {[sin(A + 60°) + sin(A - 60°) - sin(A + 60°) - sin(60° - A)] /2} = 0

(applying the trig identity sin(- α) = sin α)

1 + 2 {sin(A - 60°) - [- sin(- 60° + A)]} /2} = 0

1 + 2 {[sin(A - 60°) + sin(A - 60°)] /2} = 0

1 + 2 {[2sin(A - 60°)] /2} = 0

simplifying to:

1 + 2sin(A - 60°) = 0

then:

2sin(A - 60°) = - 1

sin(A - 60°) = - 1/2

hence:

I) (recalling that sin(- 30°) = - 1/2):

A - 60° = - 30°

A = 60° - 30° = 30°

II) (recalling that sin(180° + 30°) = - 1/2)

A - 60° = 180° + 30°

A = 60° + 180° + 30° = 180° + 90° = 270° (nevertheless this value is outside the interval (0, 90°) )

in conclusion, the only solution for 0 < A < 90° is:

A = 30°

I hope it's helpful