Question

If `sqrt(3) tan θ =1` then evaluate (cos²θ – sin²θ).

Answer 1

In right ΔABC, let ∠B = 90° and ∠BAC = θ.

Then, `sqrt(3) tan θ = 1` ⇒ `tan θ = 1/sqrt(3) = (BC)/(AB)`.

Let BC = x and `AB = sqrt(3)x`.

Then, AC² = AB² + BC²

= (3x² + x²)

= 4x²

⇒ `AC = sqrt(4x^2)`

⇒ AC = 2x

∴ `sin θ = (BC)/(AC) = x/(2x) = 1/2` and `cos θ = (AB)/(AC) = (sqrt(3)x)/(2x) = sqrt(3)/2`.

∴ `(cos^2θ - sin^2θ) = (3/4 - 1/4)`

= `2/4`

= `1/2`