Q103 / 180 · COMEDK 2023 (Memory Based) · Mathematics
The expression $\frac{2 \tan A}{1-\cot A}+\frac{2 \cot A}{1-\tan A}$ can be written as
- A$\sin 2 A+\cos 2 A$
- Correct answerB$2 \sec A \operatorname{cosec} A+2$
- C$\tan 2 A+\cot 2 A$
- D$\sec 2 A+\operatorname{cosec} 2 A$
Solution
$\begin{aligned} & \text { Given, } \frac{2 \tan A}{1-\cot A}+\frac{2 \cot A}{1-\tan A} \\ & =\frac{\frac{2 \sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{2 \cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} \\ & =\left(\frac{\sin A}{\cos A}\right)\left(\frac{2 \sin A}{\sin A-\cos A}\right)+\left(\frac{\cos A}{\sin A}\right)\left(\frac{2 \cos A}{\cos A-\sin A}\right) \\ & =\frac{2}{\sin A-\cos A}\left[\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A}\right] \\ & =\frac{2}{\sin A-\cos A}\left[\frac{\sin ^3 A-\cos ^3 A}{\sin A \cos A}\right] \\ & =\frac{2(\sin A-\cos A)\left(\sin 2 A+\cos ^2 A+\sin A \cos A\right)}{(\sin A-\cos A)(\sin A \cos A)} \\ & =2\left(\frac{1}{\sin A \cos A}+1\right) \\ & =2(\sec A \operatorname{cosec} A+1)=2 \sec A \operatorname{cosec} A+2\end{aligned}$
Hence, the correct answer is option: 'B'.