Question

Solve the equation: tan θ + sec θ = √3, 0 ≤ θ ≤ 2π

Answer 1

Answer:

The Answer is     θ = π/6    &    θ = 11π/6.

Explanation:

Given that

tan θ + sec θ = √3,        0 ≤ θ ≤ 2π

sec θ + tan θ = √3   ........... (i)

We know that

sec^2 θ - tan^2 θ = 1

(sec θ + tan θ)(sec θ - tan θ) = 1

(√3  )(sec θ - tan θ) = 1

sec θ - tan θ = 1/√3    ................(ii)

Adding (i) & (ii), we get

2 sec θ = √3 + 1/√3

2 sec θ = (3 + 1)/√3

2 sec θ = 4/√3

sec θ = 4/2√3

sec θ = 2/√3

cos θ = √3/2

The Reference Angle is π/6

Cos is +ve in Quadrant I & IV.

For Quadrant I  :  θ = π/6

For Quadrant IV :  θ =  2π - π/6 = 11π/6

So the Answer is  π/6 & 11π/6.