VerifiedHint: The above question is related to inverse trigonometric function and for solving the problem, you need to use the property that $ {{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} $ . After that use the trigonometric table to put the value of $ \cot \dfrac{\pi }{2} $ which is equal to 0.
Complete step-by-step answer:
Before starting with the solution to the above question, we will first talk about the required details of different inverse trigonometric ratios. So, we must remember that inverse trigonometric ratios are completely different from trigonometric ratios and have many constraints related to their range and domain. So, to understand these constraints and the behaviour of inverse trigonometric functions, let us look at some of the important graphs. First, let us see the graph of $ {{\tan }^{-1}}x $ .
Also, we will draw the graph of $ {{\cot }^{-1}}x $ as well.
Some of the results we can draw from the graphs is: both $ {{\cot }^{-1}}x $ and $ {{\tan }^{-1}}x $ is defined for all real values of x. Also, we can see that the range of $ {{\cot }^{-1}}x $ is $ \left( 0,\pi \right) $ and $ {{\tan }^{-1}}x $ is $ \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) $ .
Now moving to the solution to the above question, we will start with the simplification of the expression given in the question.
$ \cot \left( {{\tan }^{-1}}a+{{\cot }^{-1}}a \right) $
Now we know that $ {{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} $ . So, our expression comes out to be:
$ \cot \left( {{\tan }^{-1}}a+{{\cot }^{-1}}a \right)=\cot \dfrac{\pi }{2} $
As $ \dfrac{\pi }{2} $ is a standard angle and we know the values of trigonometric ratios for all the standard angles from the trigonometric table, we can say $ \cot \dfrac{\pi }{2}=0 $ .
$ \cot \left( {{\tan }^{-1}}a+{{\cot }^{-1}}a \right)=\cot \dfrac{\pi }{2}=0 $
Hence, the answer to the above question is option (a).
Note: It is important that you learn the complete trigonometric table, as the values of different trigonometric ratios are used very often.
It will also be helpful if you remember the graphs along with the domains and ranges of different inverse trigonometric functions, as it is seen very often that students commit mistakes in questions where the domains and range of inverse trigonometric functions are used because they are very confusing.
