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

Hint: 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.