For the given function $ f\left( x \right) = {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} $ if $ x \ne a $ and $ f\left( x \right) = k $ if $ x = a $ and $ f\left( x \right) $ is continuous at $ x = a $ then $ k = $
(A) $ {e^{\tan a}} $
(B) \[{e^{\cot a}}\]
(C) $ {e^a} $
(D) $ {e^{1/a}} $

Hint: This question is based on Limits and Derivatives. The condition of the continuity of a function is that the limit of the function should exist. To solve this question, we have to arrange this expression in such a way that it satisfies the condition of the continuity and then substitute the limit value of the function and find the value of $ k $ .

Complete step-by-step answer:
Given:
 $ f\left( x \right) = {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} $ if $ x \ne a $ and,
 $ f\left( x \right) = k $ if $ x = a $
Also, it is given that $ f\left( x \right) $ is continuous at $ x = a $ . It means that-
 $ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = k $
If we substitute $ x = a $ in this equation we get,
 $
f\left( a \right) = {\left( {\dfrac{{\sin a}}{{\sin a}}} \right)^{\dfrac{1}{{a - a}}}}\\
f\left( a \right) = {1^\infty }
 $
We cannot solve this expression by simple substitution. So, we have to write this expression in such a way that it can be solved easily. Now we know that if any expression reduces to $ {1^\infty } $ form we can use this formula –
 $ \mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}} $
So now, substituting the values into this formula we have –
 $
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\mathop {\lim }\limits_{x \to a} \dfrac{1}{{x - a}}\left( {\dfrac{{\sin x}}{{\sin a}} - 1} \right)}}\\
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\mathop {\lim }\limits_{x \to a} \dfrac{1}{{x - a}}\left( {\dfrac{{\sin x - \sin a}}{{\sin a}}} \right)}}
 $
We know the formula for $ \left( {\sin x - \sin a} \right) = 2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right) $
So, substituting this formula in the equation we get,
 $
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{1}{{\sin a}}\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{x - a}}} \right)}}\\
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{1}{{\sin a}}\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{\left( {\dfrac{{x - a}}{2}} \right)}}} \right)}}
 $
Now we know the rule that
 $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 $
So now we have to change the limit, then assume $ x - a = t{\rm{ (suppose)}} $
The limit $ x \to a $ changes to $ x - a \to 0 $ or $ t \to 0 $
Now we can apply this rule in the expression above. So, we have-
 $
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{{\cos \left( {\dfrac{{a + a}}{2}} \right)}}{{\sin a}}\left( {\mathop {\lim }\limits_{t \to 0} \dfrac{{\sin \left( {\dfrac{t}{2}} \right)}}{{\left( {\dfrac{t}{2}} \right)}}} \right)}}\\
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{{\cos a}}{{\sin a}}\left( {\mathop {\lim }\limits_{t \to 0} \dfrac{{\sin \left( {\dfrac{t}{2}} \right)}}{{\left( {\dfrac{t}{2}} \right)}}} \right)}}\\
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\cot a \times 1}}\\
\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\cot a}}
 $
Therefore, the value of $ k $ is $ {e^{\cot a}} $
So, the correct answer is “Option B”.

Note: It should be noted that for a function to be continuous the limit of the function should exist and the value of the Right-hand limit should be equal to the Left-hand limit.
For example, if a function is continuous at point $ x = a $ then
 $ \mathop {\lim }\limits_{x \to a} f\left( x \right) $ exists and
 $ \mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right) $