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

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

& \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{aligned}$$ Similarly, taking the second term we get $$\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{aligned} & \Rightarrow A=-\dfrac{\pi }{2}+2\pi \\\ & \Rightarrow A=\dfrac{3\pi }{2} \\\ \end{aligned}$$ **Therefore, the answer of the given question is $$\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.