Question

If ⊝ is an acute angle and $\frac{Cos⊝+Sin⊝}{Cos⊝-Sin⊝}$ = $\frac{1+√3}{1-√3}$, Find the value of $2 sec²⊝-3cosec²⊝$.​

Answer 1

EXPLANATION.

⇒ θ is the acute angle,

⇒ cosθ + sinθ/cosθ - sinθ = 1 + √3/1 - √3.

As we know that,

In this type of question we can use componendo and dividendo, we get.

⇒ [cosθ + sinθ] + [cosθ - sinθ]/[cosθ + sinθ] - [cosθ - sinθ] = [1 + √3] + [1 - √3]/[1 + √3] - [1 - √3].

⇒ cosθ + sinθ + cosθ - sinθ/cosθ + sinθ - cosθ + sinθ = 1 + √3 + 1 - √3/1 + √3 - 1 + √3.

⇒ 2cosθ/2sinθ = 2/2√3.

⇒ cosθ/sinθ = 1/√3.

⇒ cotθ = 1/√3.

⇒ cotθ = cot60°.

⇒ θ = 60°.

To find :

⇒ 2sec²θ - 3cosec²θ.

Put the value of θ = 60° in equation, we get.

⇒ 2sec²(60°) - 3cosec²(60°).

⇒ 2 x 2 x 2 - 3 x 2/√3 x 2/√3.

⇒ 8 - 4 = 4.

⇒ 2sec²θ - 3cosec²θ = 4.

                                                                                                                       

MORE INFORMATION.

Trigonometric ratios of multiple angles.

(1) = sin2θ = 2sinθ.cosθ = 2tanθ/1 + tan²θ.

(2) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) = tan2θ = 2tanθ/1 - tan²θ.

(4) = sin3θ = 3sinθ - 4sin³θ.

(5) = cos3θ = 4cos³θ - 3cosθ.

(6) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.