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Question: If an inverse trigonometric equation is given as \({{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}\)...

If an inverse trigonometric equation is given as tan⁡−1x+tan⁡−1y=π4{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4} , xy<1 . Then write the value of x+y+xyx+y+xy.

Explanation

Solution

As it is given that xy<1 and the looking at the form of the equation, it is very clear that you have to use the formula tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} . After that you need to take tangent of both sides of the equation followed by use of the identity tan⁡(tan⁡−1x)=x\tan \left( {{\tan }^{-1}}x \right)=x .

Complete step-by-step answer:
Before starting with the solution to the above question, we will first talk about the required details of different inverse trigonometric ratios. So, we must remember that inverse trigonometric ratios are completely different from trigonometric ratios and have many constraints related to their range and domain. So, to understand these constraints and the behaviour of inverse trigonometric functions, let us look at some of the important graphs. First, let us see the graph of sin⁡−1x{{\sin }^{-1}}x .

Now let us draw the graph of cos−1xco{{s}^{-1}}x .

Also, we will draw the graph of tan⁡−1x{{\tan }^{-1}}x as well.

So, looking at the above graphs, we can draw the conclusion that tan⁡−1x{{\tan }^{-1}}x is defined for all real values of x, i.e., the domain of the function tan⁡−1x{{\tan }^{-1}}x is all real numbers while its range comes out to be (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) . Unlike tan⁡−1x{{\tan }^{-1}}x the functions sin−1x and cos−1xsi{{n}^{-1}}x\text{ and co}{{\text{s}}^{-1}}x have the is defined only for x∈[−1,1]x\in [-1,1] .
Now moving to the solution to the above question, we will start with the simplification of the equation given in the question.
tan⁡−1x+tan⁡−1y=π4{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}
Now, we know xy<1xy<1 . So, if we use the formula tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} , we get
tan⁡−1x+y1−xy=π4{{\tan }^{-1}}\dfrac{x+y}{1-xy}=\dfrac{\pi }{4}
Now we will take the tangent of both the sides of the equations. On doing so, we get
tan(tan⁡−1x+y1−xy)=tan⁡π4tan\left( {{\tan }^{-1}}\dfrac{x+y}{1-xy} \right)=\tan \dfrac{\pi }{4}
Now we know that tan⁡(tan⁡−1A)=A\tan \left( {{\tan }^{-1}}A \right)=A and the value of tan⁡π4\tan \dfrac{\pi }{4} is equal to 1. So, we get
x+y1−xy=1\dfrac{x+y}{1-xy}=1
⇒x+y=1−xy\Rightarrow x+y=1-xy
⇒x+y+xy=1\Rightarrow x+y+xy=1
Therefore, the value of x+y+xy is equal to 1.

Note: While dealing with inverse trigonometric functions, it is preferred to know about the domains and ranges of the different inverse trigonometric functions. For example: the domain of sin⁡−1x{{\sin }^{-1}}x is [−1,1][-1,1] and the range is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] . Also, the formula tan⁡−1A+tan⁡−1B{{\tan }^{-1}}A+{{\tan }^{-1}}B is different for different cases and can be represented as:
{{\tan }^{-1}}A+{{\tan }^{-1}}B=\left\\{ \begin{aligned} & {{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{; AB < 1} \\\ & \pi +{{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{; AB > 1} \\\ \end{aligned} \right.