Question

$\left(\frac{\sin \theta }{1-\cot \theta }\right)+\left(\frac{\cos \theta }{1-\tan \theta }\right)=$

• A. $0$
• B. $1$
• C. $\cos \theta –\sin \theta$
• D. $\cos \theta +\sin \theta$

Answer 1

The correct option is D

$\cos \theta +\sin \theta$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \left(\frac{\sin \theta }{1-\cot \theta }\right)+\left(\frac{\cos \theta }{1-\tan \theta }\right) \\ =\left(\frac{\sin \theta }{1-\frac{\cos \theta }{\sin \theta }}\right)+\left(\frac{\cos \theta }{1-\frac{\sin \theta }{\cos \theta }}\right)\left[\because \cot \theta =\frac{\cos \theta }{\sin \theta }and\tan \theta =\frac{\sin \theta }{\cos \theta }\right] \\ =\left(\frac{{\sin }^2\theta }{\sin \theta –\cos \theta }\right)+\left(\frac{{\cos }^2\theta }{\cos \theta –\sin \theta }\right) \\ =\left(\frac{{\sin }^2\theta }{\sin \theta –\cos \theta }\right)-\left(\frac{{\cos }^2\theta }{\sin \theta -\cos \theta }\right)\left[\text{making the denominator to be common}\right] \\ =\frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta –\cos \theta } \\ =\frac{\left(\sin \theta –\cos \theta \right)\left(\sin \theta +\cos \theta \right)}{\left(\sin \theta –\cos \theta \right)}\left[\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right] \\ =\sin \theta +\cos \theta \end{gathered}$$

Therefore, $\left(\frac{\sin \theta }{1-\cot \theta }\right)+\left(\frac{\cos \theta }{1-\tan \theta }\right)=$$\cos \theta +\sin \theta$

Hence, option (D) is the correct answer.