A) \[\frac{1+\sin \theta }{\cos \theta }\]
B) \[\frac{1-\sin \theta }{\cos \theta }\]
C) \[\frac{\cos \theta }{1+\sin \theta }\]
D) \[\frac{1+\sin \theta }{2\cos \theta }\]
Correct Answer: A
Solution :
\[\frac{\tan \theta +sec\theta -1}{\tan \theta -\sec \theta +1}\] \[=\frac{(\tan \theta +\sec \theta )-1}{\tan \theta -\sec \theta +1}\] \[=\frac{(\sec \theta +tan\theta )-(se{{c}^{2}}\theta -{{\tan }^{2}}\theta )}{\tan \theta -\sec \theta +1}\] \[[\because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1]\] \[=\frac{\begin{align} & (\sec \theta +\tan \theta )-(\sec \theta +\tan \theta ) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\sec \theta -\tan \theta ) \\ \end{align}}{\tan \theta -\sec \theta +1}\] \[=\frac{(\sec \theta +\tan \theta )[1-\sec \theta +\tan \theta )]}{\tan \theta -\sec \theta +1}\] \[=\frac{(\sec \theta +tan\theta )(tan\theta -sec\theta +1)}{(\tan \theta -\sec \theta +1)}\] \[=\sec \theta +\tan \theta \] \[=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }\]
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