2022-M2-02 – Solving Master

Ans: (b)

\frac{5 π}{12}

$\begin{array}{cl} \frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} \\ = & \frac{\tan θ}{1 - \frac{1}{\tan θ}} + \frac{\frac{1}{\tan θ}}{1 - \tan θ} \\ = & \frac{\tan^{2} θ}{\tan θ - 1} + \frac{1}{\tan θ (1 - \tan θ)} \\ = & \frac{\tan^{3} θ - 1}{\tan θ (\tan θ - 1)} \\ = & \frac{(\tan θ - 1) (\tan^{2} θ + \tan θ + 1)}{\tan θ (\tan θ - 1)} \\ = & \frac{\tan^{2} θ + 1 + \tan θ}{\tan θ} \\ = & \frac{\sec^{2} θ + \tan θ}{\tan θ} \\ = & \frac{1}{\cos^{2} θ} \div \frac{\sin θ}{\cos θ} + 1 \\ = & 1 + \frac{1}{\cos θ \sin θ} \\ = & 1 + \sec θ \csc θ \end{array}$
$\begin{array}{rcl} \frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} & = & 5 \\ 1 + \sec θ \csc θ & = & 5, by the result of (a). \\ \frac{1}{\cos θ \sin θ} & = & 4 \\ \cos θ \sin θ & = & \frac{1}{4} \\ \frac{1}{2} \sin 2 θ & = & \frac{1}{4} \\ \sin 2 θ & = & \frac{1}{2} \end{array}$

$∴ 2 θ = \frac{π}{6}$ or $2 θ = \frac{5 π}{6}$ .

Hence, we have $θ = \frac{π}{12}$ (rejected) or $θ = \frac{5 π}{12}$ .