Q44 / 125 · BITSAT 2011 · Mathematics
\(\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}\) is equal to
- A\(\sin \theta-\cos \theta\)
- Correct answerB\(\sin \theta+\cos \theta\)
- C\(\tan \theta+\cot \theta\)
- D\(\tan \theta-\cot \theta\)
Solution
\(\begin{aligned}
& \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \\
= & \frac{\cos ^2 \theta}{\cos \theta-\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta
\end{aligned}\)
Hence, the correct answer is option: 'B'.