Question

If $\sec \dfrac{\left( {{x}^{2}}-{{y}^{2}} \right)}{\left( {{x}^{2}}+{{y}^{2}} \right)}={{e}^{a}}$ , then $\dfrac{dy}{dx}$ is equal to
A. $\dfrac{{{y}^{2}}}{{{x}^{2}}}$
B. $\dfrac{y}{x}$
C. $\dfrac{x}{y}$
D. $\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}$

Answer 1

For such types of questions always take trigonometric functions to the other side to make them in the form of inverse trigonometric functions. Then take all other variables other than ‘y’ to the other side as we want to find out the value of $\dfrac{dy}{dx}$ , and differentiate it, so we will get the answer.

Complete step by step answer:
Moving ahead with the question in step wise manner;
As we want to find out the $\dfrac{dy}{dx}$ , means we need to find out the differentiation of ‘y’ in terms of ‘x’. Now as we have question $\sec \dfrac{\left( {{x}^{2}}-{{y}^{2}} \right)}{\left( {{x}^{2}}+{{y}^{2}} \right)}={{e}^{a}}$ , as in sec we have angle in the terms of ‘x’ and ‘y’ both whose differentiation is difficult for us to do. So let us take the sec to RHS side which will make it in inverse form, moreover ${{e}^{a}}$ is some constant so trigonometric inverse of some constant is also constant i.e.
$\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}={{\sec }^{-1}}\left( {{e}^{a}} \right)$ by applying componendo and dividendo in the equation we will get;
$\dfrac{2{{x}^{2}}}{-2{{y}^{2}}}=\dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}$
On further solving it we will get;
$\begin{aligned} & \dfrac{{{x}^{2}}}{-{{y}^{2}}}=\dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1} \\\ & \dfrac{-{{y}^{2}}}{{{x}^{2}}}=\dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \\\ & {{y}^{2}}=-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right){{x}^{2}} \\\ \end{aligned}$
Now as we know that ${{e}^{a}}$ is constant, then its inverse trigonometric will also be constant so we can say that ${{\sec }^{-1}}\left( {{e}^{a}} \right)$ is also constant, add any mathematical operation with constant is always constant so we can say that $\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)$ is constant.
As now to find the $\dfrac{dy}{dx}$ directly we can differentiate the equation, as in LHS side we will get $2y\dfrac{dy}{dx}$ and in RHS we will get $-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)2x$ , as $\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)$ is constant so there will no differentiation of it, so we will get;
$\begin{aligned} & 2y\dfrac{dy}{dx}=-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)2x \\\ & \dfrac{dy}{dx}=-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)\dfrac{2x}{2y} \\\ & \dfrac{dy}{dx}=-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)\dfrac{x}{y} \\\ \end{aligned}$
So we got $\dfrac{dy}{dx}=-\left( \dfrac{{{\sec }^{-1}}\left( {{e}^{a}} \right)-1}{{{\sec }^{-1}}\left( {{e}^{a}} \right)+1} \right)\dfrac{x}{y}$ , but option are in the form of only ‘x’ and ‘y’ so now we had to remove the inverse trigonometric function, so from above solution we got $\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}={{\sec }^{-1}}\left( {{e}^{a}} \right)$ , so from here we have the relation between inverse trigonometric function and ‘x’ and ‘y’ so replace them, which will give us;
$\dfrac{dy}{dx}=-\left( \dfrac{\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}-1}{\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}+1} \right)\dfrac{x}{y}$
On further simplifying we will get;
$\begin{aligned} & \dfrac{dy}{dx}=-\left( \dfrac{\dfrac{{{x}^{2}}-{{y}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}}{\dfrac{{{x}^{2}}-{{y}^{2}}+\left( {{x}^{2}}+{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}} \right)\dfrac{x}{y} \\\ & \dfrac{dy}{dx}=-\left( \dfrac{-2{{y}^{2}}}{2{{x}^{2}}} \right)\dfrac{x}{y} \\\ & \dfrac{dy}{dx}=\dfrac{y}{x} \\\ \end{aligned}$
So we got $\dfrac{dy}{dx}=\dfrac{y}{x}$ .
Hence answer is $\dfrac{dy}{dx}=\dfrac{y}{x}$

So, the correct answer is “Option B”.

Note: We can say that taking the trigonometric function to the RHS side to simplify the function for solving is one of the types of dealing with the trigonometric function, which you can do any time with similar types of questions.