Question : $\frac{1+\sin \theta}{\cos \theta}$ is equal to which of the following (where $\left.\theta \neq \frac{\pi}{2}\right)?$
Option 1: $\frac{1+\cos \theta}{\sin \theta}$
Option 2: $\frac{\tan \theta+1}{\tan \theta-1}$
Option 3: $\frac{\tan \theta-1}{\tan \theta+1}$
Option 4: $\frac{\cos \theta}{1-\sin \theta}$
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Correct Answer: $\frac{\cos \theta}{1-\sin \theta}$
Solution : Given, $\frac{1+\sin \theta}{\cos \theta}$ Multiplying numerator and denominator by $(1-\sin \theta)$ = $\frac{(1+\sin \theta)\times (1-\sin \theta)}{\cos \theta\times (1-\sin \theta)}$ = $\frac{(1-\sin^2 \theta)}{\cos \theta\times (1-\sin \theta)}$ = $\frac{\cos^2 \theta}{\cos \theta\times (1-\sin \theta)}$ = $\frac{\cos \theta}{1-\sin \theta}$ Hence, the correct answer is $\frac{\cos \theta}{1-\sin \theta}$.
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