If (cos²θ - 1)(2sec²θ) + sec²θ + 2tan²θ = 2, 0° (secθ + sinθ)over (cosecθ - cosθ) is equal to:

If (cos²θ - 1)(2sec²θ) + sec²θ + 2tan²θ = 2, 0° (secθ +…

If (cos²θ - 1)(2sec²θ) + sec²θ + 2tan²θ = 2, 0° < θ < 90°, then $(secθ + sinθ)\over (cosecθ - cosθ)$ is equal to:

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Correct answer: D

Explanation

The correct answer is Option 4 i.e. 3.

(cos²θ - 1)(2sec²θ) + sec²θ + 2tan²θ = 2

⇒ (-sin²θ).2sec²θ + sec²θ + 2tan²θ = 2

⇒ -2tan²θ + 1 + tan²θ + 2tan²θ = 2

⇒ tan²θ = 1

⇒ θ = 45°

Now, (secθ + sinθ)/(cosecθ - cosθ)

= (sec45° + sin45°)/(cosec45° - cos45°)

= (√2 + 1/√2)/(√2 - 1/√2)

= (3/√2)/(1/√2)

= 3