Question

If ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$ and both $x$ and $y$ are positive and integral, then $x$ and $y$ are equal to

• A. $(1,2)$ and $(2,7)$
• B. $(1,2)$ and $(1,7)$
• C. $(1,7)$ and $(2,7)$
• D. $(1,2)$ and $(2,1)$

Answer 1

The correct option is A

$(1,2)$ and $(2,7)$

Explanation for the correct option:

Step 1: Simplify the given relation ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}..........\left(i\right)$ then find $x$ and $y$ .

Let us assume $m={\sin }^{-1}\frac{3}{\sqrt{10}}$ and $n={\cos }^{-1}\frac{y}{\sqrt{1+y^2}}$

which implies $\sin m=\frac{3}{\sqrt{10}}$ and $\cos n=\frac{y}{\sqrt{1+y^2}}$, then

$\begin{array}{c}\cos m=\sqrt{1-{\sin }^{2}m}\\ =\sqrt{1-\frac{9}{10}}\\ =\sqrt{\frac{10-9}{10}}\\ =\frac{1}{\sqrt{10}}\end{array}$

Therefore,

$\begin{array}{c}\tan m=\frac{\sin m}{\cos m}\\ =\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}\\ =3\end{array}$

Which implies $m={\tan }^{-1}3$ and similarly,

$\begin{array}{c}\sin n=\sqrt{1-{\cos }^{2}n}\\ =\sqrt{1-\frac{y^2}{1+y^2}}\\ =\sqrt{\frac{1+y^2-y^2}{1+y^2}}\\ =\frac{1}{\sqrt{1+y^2}}\end{array}$

Therefore,

$\begin{array}{c}\tan n=\frac{\sin n}{\cos n}\\ =\frac{\frac{1}{\sqrt{1+y^2}}}{\frac{y}{\sqrt{1+y^2}}}\\ =\frac{1}{y}\end{array}$

Which implies $n={\tan }^{-1}\frac{1}{y}$.

Step 2: Use the formula ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{{\displaystyle A-B}}{{\displaystyle 1+AB}}\right)$ after substituting $m={\tan }^{-1}3$ and $n={\tan }^{-1}\frac{1}{y}$ in the given equation,

$\begin{array}{c}{\tan }^{-1}x+{\tan }^{-1}\frac{1}{y}={\tan }^{-1}3\\ {\tan }^{-1}\frac{1}{y}={\tan }^{-1}3-{\tan }^{-1}x\\ {\tan }^{-1}\frac{1}{y}={\tan }^{-1}\left(\frac{3-x}{1+3x}\right)\\ \frac{1}{y}=\frac{3-x}{1+3x}\\ 1+3x=3y-xy\\ 1+xy=3(y-x)..............\left(ii\right)\end{array}$

Step 3: Put $x=1,2$ in the equation (ii), as in given the option only these values of $x$ is available then find $y$

when $x=1$, then

$\begin{array}{c}1+y=3(y-1)\\ 1+y=3y-3\\ 2y=4\\ y=2\end{array}$

again, when $x=2$, then

$\begin{array}{c}1+2y=3(y-2)\\ 1+2y=3y-6\\ y=7\end{array}$

Therefore, $x$ and $y$ are equal to $(1,2)$ and $(2,7)$.

Hence, the correct option is (A).