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Question

Mathematics Question on Trigonometric Functions

If cosecθ−cot⁡ θ=qcosec \theta - \cot \, \theta = q, then the value of cot⁡ θ\cot \, \theta is:

A

1−q22q\frac{1 - q^2}{2q}

B

1+q2q\frac{1 + q^2}{q}

C

q1−q2\frac{q}{1 - q^2}

D

q1+q2\frac{q}{1 + q^2}

Answer

1−q22q\frac{1 - q^2}{2q}

Explanation

Solution

Given : cosec θ−cot⁡ θ=qcosec \, \theta - \cot \, \theta = q We know that 1+cot⁡2θ=cosec2θ1 + \cot^2 \theta = cosec^2 \theta ∴ 1+cot⁡2θ−cot⁡θ=q\therefore \, \sqrt{1 + \cot^2 \theta} - \cot \theta = q ⇒  1+cot⁡2θ=q+cot⁡θ\Rightarrow \, \, \sqrt{1 + \cot^2 \theta} = q + \cot \theta Squaring both sides, we get ∴ (1+cot⁡2θ)=(q+cot⁡θ)2\therefore \, \left(1+ \cot^{2} \theta\right) = \left(q + \cot \theta\right)^{2} ⇒1+cot⁡2θ=q2+cot⁡2θ+2qcot⁡θ\Rightarrow 1 + \cot^{2} \theta = q^{2} + \cot^{2} \theta+ 2 q \cot\theta ⇒1−q2=2qcot⁡θ\Rightarrow 1 - q^{2} = 2q \cot\theta ⇒cot⁡θ=1−q22q\Rightarrow \cot \theta = \frac{ 1- q^{2}}{2q}