Question

Find the value of \[-{{\sin }^{-1}}\left( \dfrac{{{\cot }^{-1}}\left( -1 \right)+{{\cot }^{-1}}\left( -2 \right)+{{\cot }^{-1}}\left( -3 \right)}{{{\tan }^{-1}}\left( -1 \right)+{{\tan }^{-1}}\left( -\dfrac{1}{2} \right)+{{\tan }^{-1}}\left( -\dfrac{1}{3} \right)} \right)+2\pi \]

Answer 1

Hint: First we need to convert all the tangent angles in the denominator to cotangent angles by using the formula\[{{\cot }^{-1}}\left( -x \right)={{\tan }^{-1}}\left( \dfrac{1}{x} \right)\]. Then we evaluate the value of \[{{\sin }^{-1}}\theta \] after converting all the tangent angles in the denominator to cotangent angles. We need use the range of \[{{\sin }^{-1}}\theta \] is
\[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]. We have to use \[\theta \] in this range only.

Complete step by step answer:
We know that tangent and cotangent are reciprocal to each other that is
\[{{\cot }^{-1}}\left( -x \right)={{\tan }^{-1}}\left( \dfrac{1}{x} \right)\]
By using the above formulae let us covert \[{{\tan }^{-1}}\left( -1 \right),{{\tan }^{-1}}\left( -\dfrac{1}{2} \right),{{\tan }^{-1}}\left( -\dfrac{1}{3} \right)\]
Taking the first term we get
\[\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( -1 \right)={{\cot }^{-1}}\left( \dfrac{1}{-1} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( -1 \right)={{\cot }^{-1}}\left( -1 \right) \\
\end{align}\]
Similarly, taking the second term we get
\[\Rightarrow {{\tan }^{-1}}\left( -\dfrac{1}{2} \right)={{\cot }^{-1}}\left( -2 \right)\]
Now taking the third term we get
\[\Rightarrow {{\tan }^{-1}}\left( -\dfrac{1}{3} \right)={{\cot }^{-1}}\left( -3 \right)\]
Now let us assume that the given question as
\[\Rightarrow A=-{{\sin }^{-1}}\left( \dfrac{{{\cot }^{-1}}\left( -1 \right)+{{\cot }^{-1}}\left( -2 \right)+{{\cot }^{-1}}\left( -3 \right)}{{{\tan }^{-1}}\left( -1 \right)+{{\tan }^{-1}}\left( -\dfrac{1}{2} \right)+{{\tan }^{-1}}\left( -\dfrac{1}{3} \right)} \right)+2\pi \]
Now let us substitute the values of \[{{\tan }^{-1}}\left( -1 \right),{{\tan }^{-1}}\left( -\dfrac{1}{2} \right),{{\tan }^{-1}}\left( -\dfrac{1}{3} \right)\] in the above equation we will get
\[\Rightarrow A=-{{\sin }^{-1}}\left( \dfrac{{{\cot }^{-1}}\left( -1 \right)+{{\cot }^{-1}}\left( -2 \right)+{{\cot }^{-1}}\left( -3 \right)}{{{\cot }^{-1}}\left( -1 \right)+{{\cot }^{-1}}\left( -2 \right)+{{\cot }^{-1}}\left( -3 \right)} \right)+2\pi \]
Here, since the numerator and denominator is equal by cancelling it then we will get
\[\Rightarrow A=-{{\sin }^{-1}}\left( 1 \right)+2\pi ............equation(i)\]
Here we know that the range of \[{{\sin }^{-1}}\theta \]is\[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]. We have to use \[\theta \] in this range only.
We also know that if \[\sin \theta =1\] then we can write
\[\theta =2n\pi +\dfrac{\pi }{2},n=0,1,2,.........\]
Here in the same way
\[{{\sin }^{-1}}\left( 1 \right)=2n\pi +\dfrac{\pi }{2},n=0,1,2,........\]
But the range of \[{{\sin }^{-1}}\theta \]is\[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\], \[n=0\] is the only option.
Therefore, \[{{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2}\]
By substituting this value in equation (i) we will get
\[\begin{align}
  & \Rightarrow A=-\dfrac{\pi }{2}+2\pi \\
 & \Rightarrow A=\dfrac{3\pi }{2} \\
\end{align}\]

Therefore, the answer of the given question is \[\dfrac{3\pi }{2}\].

Note: Some students will make mistakes in taking the values of \[{{\sin }^{-1}}\theta \]. If the same question is asked in multiple choice questions then students will take every value of ‘n’ in\[{{\sin }^{-1}}\left( 1 \right)=2n\pi +\dfrac{\pi }{2},n=0,1,2,.........\]and gives all possible answers but that is very wrong. We need to take the values in such a way that they lie inside the range of a function. A function cannot give the values of outside of its range.