?
$$\sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} $$ is equal to-
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} \cr
& = \sqrt {\frac{{\cos \theta \left( {\frac{1}{{\sin \theta }} + 1} \right)}}{{\cos \theta \left( {\frac{1}{{\sin \theta }} - 1} \right)}}} \cr
& = \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \cr
& = \sqrt {\frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \cr
& = \frac{{1 + \sin \theta }}{{\cos \theta }} \cr
& = \sec \theta + \tan \theta \cr} $$
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