Question

The no. of positive integral solutions $ \left( x,y \right) $ of the equation $ {{\tan }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)={{\tan }^{-1}}\left( 3 \right) $ are?

Answer 1

Hint: We start solving the problem by using the fact $ {{\cot }^{-1}}\left( a \right)={{\tan }^{-1}}\left( \dfrac{1}{a} \right) $ in the given trigonometric equation. We then make the necessary arrangements and make use of the fact $ {{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right) $ to proceed through the problem. We then make the necessary calculations to get the relation between x and y. We then check the values of x and y for which the given conditions are satisfied to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the total no. of positive integral solutions $ \left( x,y \right) $ of the equation $ {{\tan }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)={{\tan }^{-1}}\left( 3 \right) $ .
We have given $ {{\tan }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)={{\tan }^{-1}}\left( 3 \right) $ .
We know that $ {{\cot }^{-1}}\left( a \right)={{\tan }^{-1}}\left( \dfrac{1}{a} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{1}{y} \right)={{\tan }^{-1}}\left( 3 \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{y} \right)={{\tan }^{-1}}\left( 3 \right)-{{\tan }^{-1}}\left( x \right) $ .
We know that $ {{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{y} \right)={{\tan }^{-1}}\left( \dfrac{3-x}{1+3x} \right) $ .
 $ \Rightarrow \dfrac{1}{y}=\dfrac{3-x}{1+3x} $ .
 $ \Rightarrow y=\dfrac{1+3x}{3-x} $ ---(1).
According to the problem, we need to find the positive integral solutions for $ \left( x,y \right) $ . From equation (1), we can see that the value of y is undefined when $ x=3 $ and negative when the $ x > 3 $ .
So, the only integer values possible for x is 1 and 2.
Now, let us find the value of y when $ x=1 $ .
So, we have $ y=\dfrac{1+3\left( 1 \right)}{3-1} $ .
 $ \Rightarrow y=\dfrac{1+3}{2} $ .
 $ \Rightarrow y=\dfrac{4}{2}=2 $ .
So, we have found one of the positive integral solutions as $ \left( 1,2 \right) $ .
Now, let us find the value of y when $ x=2 $ .
So, we have $ y=\dfrac{1+3\left( 2 \right)}{3-2} $ .
 $ \Rightarrow y=\dfrac{1+6}{1} $ .
 $ \Rightarrow y=7 $ .
So, we have found one of the positive integral solutions as $ \left( 2,7 \right) $ .
We have found the number of positive integral solutions as 2.
 $ \therefore $ The no. of positive integral solutions $ \left( x,y \right) $ of the equation $ {{\tan }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)={{\tan }^{-1}}\left( 3 \right) $ is 2.

Note:
 We should keep in mind that infinity will not be included in the solution set for y as the given equation will not be satisfied on substituting it. We can also make use of the fact $ {{\cot }^{-1}}\left( y \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( y \right) $ for $ y > 0 $ instead of $ {{\cot }^{-1}}\left( a \right)={{\tan }^{-1}}\left( \dfrac{1}{a} \right) $ to solve the problem. We should not make calculation mistakes while solving this problem. Similarly, we can expect problems to find the integral solutions $ \left( x,y \right) $ for $ {{\left( {{\sin }^{-1}}\left( x \right) \right)}^{2}}+{{\left( {{\cos }^{-1}}\left( y \right) \right)}^{2}}=\dfrac{{{\pi }^{2}}}{4} $ .