Ans: (b) $\dfrac{5\pi}{12}$
-
$\begin{array}{cl}
& \dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta} \\
= & \dfrac{\tan\theta}{1-\frac{1}{\tan\theta}}+\dfrac{\frac{1}{\tan\theta}}{1-\tan\theta} \\
= & \dfrac{\tan^2\theta}{\tan\theta-1}+\dfrac{1}{\tan\theta(1-\tan\theta)} \\
= & \dfrac{\tan^3\theta-1}{\tan\theta(\tan\theta -1)} \\
= & \dfrac{(\tan\theta-1)(\tan^2\theta+\tan\theta+1)}{\tan\theta(\tan\theta -1)} \\
= & \dfrac{\tan^2\theta+1+\tan\theta}{\tan\theta} \\
= & \dfrac{\sec^2\theta+\tan\theta}{\tan\theta} \\
= & \dfrac{1}{\cos^2\theta}\div\dfrac{\sin\theta}{\cos\theta}+1 \\
= & 1+\dfrac{1}{\cos\theta\sin\theta} \\
= & 1+\sec\theta \csc\theta
\end{array}$ -
$\begin{array}{rcl}
\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta} & = & 5 \\
1+\sec\theta\csc\theta & = & 5 \text{ , by the result of (a).} \\
\dfrac{1}{\cos\theta\sin\theta} & = & 4 \\
\cos\theta\sin\theta & = & \dfrac{1}{4} \\
\dfrac{1}{2}\sin2\theta & = & \dfrac{1}{4} \\
\sin 2\theta & = & \dfrac{1}{2}
\end{array}$$\therefore 2\theta=\dfrac{\pi}{6}$ or $2\theta=\dfrac{5\pi}{6}$.
Hence, we have $\theta=\dfrac{\pi}{12}$ (rejected) or $\theta=\dfrac{5\pi}{12}$.
