If cotθ = 1/(√3) , 0° (2-sin²θ)/(1-cos²θ) + (cosec²θ - secθ) is:

If cotθ = 1/(√3) , 0° (2-sin²θ)/(1-cos²θ) + (cosec²θ…

If cotθ = $\frac{1}{√3}$ , 0° < θ° < 90°, then the value of $\frac{2-sin^2\theta}{1-cos^2\theta}$ + (cosec²θ - secθ) is:

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Correct answer: C

Explanation

The correct answer is Option 3 i.e. 1.

cotθ = $\frac{1}{√3}$

cotθ = cot 60°

θ = 60°

Then,

The value of $\frac{2-sin^2\theta}{1-cos^2\theta}$ + (cosec²θ - secθ)

$\frac{2-sin^260°}{1-cos^260°}$ + (cosec²60° - sec60°)

⇒ $\frac{2-({\sqrt3\over2})^2}{1-({1\over2})^2}$ + ( $({2\over{\sqrt3}})^2$ - $(2)$ )

⇒ $\frac{2-{3\over4}}{1-{1\over4}}$ + ( ${4\over{3}}$ - 2)

⇒ $\frac{8-3}{4-1}$ + (4 - 6)/3

⇒ $\frac{5}{3}$ + ${-2\over3}$ = 3/3 = 1