Question

If $\tan (x+y)+\tan (x-y)=1$ , then $\dfrac{dy}{dx}=$
A. $\dfrac{[{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)]}{[{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)]}$
B. $\dfrac{[{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)]}{[{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x+y)]}$
C. $\dfrac{[{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)]}{[{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)]}$
D. None of these

Answer 1

We have to apply the differentiation formula in the given equation then we will find out the differentiation by applying chain rule after that take $\dfrac{dy}{dx}$ in common from the obtained equation then find out the value of $\dfrac{dy}{dx}$ and check which option is correct in the given options.

Complete step by step answer:
Let $x$ be a variable quantity, after some increment its value ${{x}_{1}}$ becomes ${{x}_{2}}$ . The difference between ${{x}_{1}}$ and ${{x}_{2}}$ is called the change in value of $x$. It is represented by $\delta x$. The change $\delta x$ may be positive or negative.
Let us suppose that $y$ is a function of $x$ and if $\delta x$ be the change in $x$ then the corresponding change in $y$ be denoted by $\delta y$.
The process of finding the differential coefficient of a function is called differentiation or we can say that differentiation is a process where we find the instantaneous rate of change in a function based on one of its variables.
The derivative or differential coefficient of a function can be obtained directly by the definition of differentiation without using addition, multiplication and quotient formulae of differentiation.
The coefficient of $y$ with respect to $x$ is represented by $\dfrac{dy}{dx}$ .
We know that the meaning of $dx$ is the increment in $x$ and it may be positive or negative. Similarly $dy$means the increment in the value of $y$
If there is an increment in the value of $x$ and $y$ in the same direction either both positive or both negative then the value of $\dfrac{dy}{dx}$ is always positive. On the other hand if the increments in $x$ and $y$ are of positive direction that is one positive and other negative then the value of $\dfrac{dy}{dx}$ will be negative.

Now according to the question:
We have given that $\tan (x+y)+\tan (x-y)=1$
Now by applying differentiation formula $\dfrac{d}{dx}\tan (x)={{\sec }^{2}}x$ we will get:
\Rightarrow $$$$\dfrac{d}{dx}[\tan (x+y)+\tan (x-y)]=\dfrac{d}{dx}(1)
Now by chain rule:
\Rightarrow $$$${{\sec }^{2}}(x+y)\cdot \dfrac{d}{dx}(x+y)+{{\sec }^{2}}(x-y)\cdot \dfrac{d}{dx}(x-y)=0
\Rightarrow $$$${{\sec }^{2}}(x+y)\cdot (1+\dfrac{dy}{dx})+{{\sec }^{2}}(x-y)\cdot (1-\dfrac{dy}{dx})=0
Multiplying the terms we will get:
\Rightarrow $$$${{\sec }^{2}}(x+y)+{{\sec }^{2}}(x+y)\dfrac{dy}{dx}+{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x-y)\dfrac{dy}{dx}=0
\Rightarrow $$$${{\sec }^{2}}(x+y)\dfrac{dy}{dx}-{{\sec }^{2}}(x-y)\dfrac{dy}{dx}+{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)=0
Now we will take $\dfrac{dy}{dx}$ in common
\Rightarrow $$$$\dfrac{dy}{dx}[{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)]+{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)=0
Now we will find out the value of $\dfrac{dy}{dx}$
\Rightarrow $$$$\dfrac{dy}{dx}[{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)]=-[{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)]
\Rightarrow $$$$\dfrac{dy}{dx}=\dfrac{-[{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)]}{[{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)]}
\Rightarrow $$$$\dfrac{dy}{dx}=\dfrac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x+y)]}

So, the correct answer is “Option 2”.

Note: Students must understand the difference between $\dfrac{\delta y}{\delta x}$ and $\dfrac{dy}{dx}$ . Here $\dfrac{\delta y}{\delta x}$ is a fraction with $\delta y$ as a numerator and $\delta x$ as a denominator while $\dfrac{dy}{dx}$ is not a fraction. It is a limiting value of $\dfrac{\delta y}{\delta x}$ . The differential coefficients of those functions which start with $'co'$ like $\text{cosx,cotx,cosecx}$ are always negative.