If $\tan (θ / 2) = cosec θ - \sin θ$ , then $c {os}^{2} (θ / 2)$ is equal to

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$\sin 36^{\circ}$

$\sin 18^{\circ} \sin 18^{\circ}$

$\cos 18^{\circ}$

$\cos 36^{\circ}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{\sin (θ / 2)}{\cos (θ / 2)} = \frac{1 - \sin^{2} θ}{\sin θ} \Rightarrow 2 \sin^{2} (θ / 2) = \cos^{2} θ \\ \Rightarrow \cos^{2} θ + \cos θ - 1 = 0 \\ \Rightarrow \cos θ = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2} \end{array}$

But $\cos θ \neq \frac{- 1 - \sqrt{5}}{2}$ so $\cos θ = \frac{\sqrt{5} - 1}{2}$

$\begin{aligned} \Rightarrow 2 \cos^{2} (θ / 2) = \frac{\sqrt{5} - 1}{2} + 1 = \frac{\sqrt{5} + 1}{2} \\ \Rightarrow \cos^{2} (θ / 2) = \frac{\sqrt{5} + 1}{4} = \cos 36^{8} . \end{aligned}$