Trigonometry help!!!!!!!!!!!!!!please??????

Question

if, cosA+secA =√3, then prove, cos^3A+sec^3A=0

Answer 1

cosA + secA = √3

cosA + 1 / cosA = √3

Multiply each side by cosA

cos²A + 1 = √3cosA

cos²A -√3cosA + 1 = 0

Substitute a = 1 , b = -√3 and c = 1 in the quadratic form

cosA = (√3 ± √3 - 4) / 2(1)

          = (√3 ± √ -1) / 2   

But -1 = i²

        =(√3 ± √i²) / 2

cosA=(√3 ± i ) / 2

secA = 1 / cosA = 2 / (√3 ± i)

                             = 2( (√3 ∓ i) / (√3 ∓ i) (√3 ± i))

                             = 2(√3 ∓ i) / (3 - i²)

Substitute i² = -1

                              = 2(√3 ∓ i) / (3 + 1)

secA                     = (√3 ∓ i) / 2

cos³A + sec³A = [(√3 ± i ) / 2]³ + [(√3 ∓ i) / 2]³

                               = 1 / 8[(√3 ± i )³ + √3 ∓ i)³]

(a ± b)³ + (a ∓ b)³ = 2a(a² + 3b²) and 2a(a² + 3b²)

Substitute a = √3 and b = i and a² = 3 and b² = -1 in the above formula

                               = 1 / 8[ 2√3(3 + 3(-1)] and 1 / 8[ 2√3(3 + 3(-1)]

Simplify

                               = 0 and 0

                               = 0.

 

Answer 2

If

Now the value of

Apply the formula

Apply the formula