Question

If ${{\cot }^{-1}}x+{{\cot }^{-1}}y=0$ and ${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$ . So $x+y=$

Answer 1

Hint: Here in this question we have been asked to find the value of $x+y$ given that ${{\cot }^{-1}}x+{{\cot }^{-1}}y=0$ and ${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$ . For answering this question we will solve both the expressions using the identities given as ${{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}$ and ${{\cot }^{-1}}x={{\cos }^{-1}}\left( \dfrac{x}{\sqrt{1+{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$ .

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $x+y$ given that ${{\cot }^{-1}}x+{{\cot }^{-1}}y=0$ and ${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$ .
From the basic concepts of inverse trigonometry, we know the following identities given as${{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}$ and ${{\cot }^{-1}}x={{\cos }^{-1}}\left( \dfrac{x}{\sqrt{1+{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$ .
We know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ .
Now we will apply $\cos $ on both sides of ${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$ then we will have
$\begin{align}
  & \Rightarrow \cos \left( {{\cos }^{-1}}x+{{\cos }^{-1}}y \right)=\cos \dfrac{\pi }{2} \\
 & \Rightarrow \cos \left( {{\cos }^{-1}}x \right)\cos \left( {{\cos }^{-1}}y \right)-\sin \left( {{\cos }^{-1}}x \right)\sin \left( {{\cos }^{-1}}y \right)=0 \\
\end{align}$
by using the formulae we have discussed above.
Now if we further simplify the expression we will get $xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=0$ .
Now we will simplify this and then we will get
$\begin{align}
  & \Rightarrow xy=\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}=1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=1 \\
\end{align}$ .
Let us assume that ${{\cot }^{-1}}x={{\theta }_{1}}$ and ${{\cot }^{-1}}y={{\theta }_{2}}$. Now from the expression ${{\cot }^{-1}}x+{{\cot }^{-1}}y=0$ we will get ${{\theta }_{1}}=-{{\theta }_{2}}$ .
As ${{\cot }^{-1}}x={{\theta }_{1}}$ we will get ${{\cot }^{-1}}x={{\cos }^{-1}}\left( \dfrac{x}{\sqrt{1+{{x}^{2}}}} \right)\Rightarrow {{\theta }_{1}}$ . Similarly ${{\cot }^{-1}}y={{\cos }^{-1}}\left( \dfrac{y}{\sqrt{1+{{y}^{2}}}} \right)\Rightarrow {{\theta }_{2}}$
Hence we can say that $\cos {{\theta }_{1}}=\cos {{\theta }_{2}}$ since ${{\theta }_{1}}=-{{\theta }_{2}}$ . So we can say that $\dfrac{x}{\sqrt{1+{{x}^{2}}}}=\dfrac{y}{\sqrt{1+{{y}^{2}}}}$ .
By simplifying this we will get
$\begin{align}
  & \Rightarrow \dfrac{{{x}^{2}}}{1+{{x}^{2}}}=\dfrac{{{y}^{2}}}{1+{{y}^{2}}} \\
 & \Rightarrow {{x}^{2}}={{y}^{2}} \\
\end{align}$
Now we will have ${{x}^{2}}+{{y}^{2}}=1$ and ${{x}^{2}}={{y}^{2}}$ . Now we need to solve these two expressions in order to get the value of $x+y$ . Now we will get $2{{x}^{2}}=1\Rightarrow {{x}^{2}}=\dfrac{1}{2}$ .
From ${{x}^{2}}={{y}^{2}}$ we will have $x=\pm y$ which gives us two different conditions $x=y$ and $x=-y$ .
When $x=y$ then we will have $x=y=\dfrac{1}{\sqrt{2}}$ and $x+y=\sqrt{2}$ .
When $x=-y$ we will have $x+y=0$ .
Therefore we can conclude that there are two possible values for $x+y$ they are 0 and $\sqrt{2}$ .

Note: In the process of answering questions of this type we should be very careful to process every possible condition. Someone may miss any one of the two conditions $x=\pm y$ and end up having a conclusion with any one answer only which would be a wrong answer.