Question

If ${{\tan }^{-1}}\dfrac{x-1}{x-2}+{{\tan }^{-1}}\dfrac{x+1}{x+2}=\dfrac{\pi }{4}$ , then find values of x.

Answer 1

Hint: At first, use the identity ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ where a represent $\dfrac{x-1}{x-2}$ and b represent as $\dfrac{x+1}{x+2}$ , then do further calculation and simplification. At last take tan on both sides and use the fact that $\tan \dfrac{\pi }{4}=1$ .

Complete step-by-step answer:

In the question, we are given a equation ${{\tan }^{-1}}\dfrac{x-1}{x-2}+{{\tan }^{-1}}\dfrac{x+1}{x+2}=\dfrac{\pi }{4}$ and we have to find the values of x for which x satisfy.
Before proceeding, let us know what are inverse trigonometric functions. Trigonometric functions are functions that are inverse function of trigonometric functions. Specifically, they are inverses of sine, cosine, tangent, cotangent, secant, cosecant functions and are used to obtain angle from any of the angle’s trigonometric ratios.
There are certain notations which are used. Some of the most common notation is using arc $\sin \left( x \right)$ , arc $\cos \left( x \right)$ , arc $\tan \left( x \right)$. These arise from geometric relationships. When measured in radians, an angle $\theta$ radian will correspond to an arc whose length is $r\theta$ , where r is radius of circle. Thus, in the unit circle, “the arc whose cosine is x” is the same as “the angle whose cosine is x”, because the length of the arc of circle in radii is the same as the measurement of angle in radians.
In the question, we are given an equation ${{\tan }^{-1}}\left( \dfrac{x-1}{x-2} \right)+{{\tan }^{-1}}\left( \dfrac{x+1}{x+2} \right)=\dfrac{\pi }{4}$ and we have to find the values of x for which x satisfy.
So, in the question we are given that,
${{\tan }^{-1}}\left( \dfrac{x-1}{x-2} \right)+{{\tan }^{-1}}\left( \dfrac{x+1}{x+2} \right)=\dfrac{\pi }{4}$
Now we will change the sum of two terms and transform it into one ratio of ${{\tan }^{-1}}$ .
We will do this using an identity which is ${{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$
Here, we will use a as $\dfrac{x-1}{x-2}$ and b as $\dfrac{x+1}{x+2}$. So, we can write it as,
${{\tan }^{-1}}\left( \dfrac{\dfrac{x-1}{x-2}+\dfrac{x+1}{x+2}}{1-\dfrac{\left( x-1 \right)}{\left( x-2 \right)}\times \dfrac{\left( x+1 \right)}{\left( x+2 \right)}} \right)=\dfrac{\pi }{4}$
Now let’s take L.C.M in the numerator and use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so we get,
${{\tan }^{-1}}\left( \dfrac{\dfrac{\left( x-1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x-2 \right)}{{{x}^{2}}-4}}{1-\dfrac{{{x}^{2}}-1}{{{x}^{2}}-4}} \right)=\dfrac{\pi }{4}$
Now, we will multiply by ${{x}^{2}}-4$ to both the numerator and also the denominator. So, we get
${{\tan }^{-1}}\left( \dfrac{\left( x-1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x-2 \right)}{{{x}^{2}}-4-\left( {{x}^{2}}-1 \right)} \right)$ which is equal to $\dfrac{\pi }{4}$ .
Now on further simplification we get,
${{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+2x-x-2+{{x}^{2}}-2x+x-2}{{{x}^{2}}-4-{{x}^{2}}+1} \right)=\dfrac{\pi }{4}$
Now on rearranging and simplifying we get,
${{\tan }^{-1}}\left( \dfrac{2{{x}^{2}}-4}{-3} \right)=\dfrac{\pi }{4}$
Now let’s take tan on both the sides. So, we get
$\dfrac{2{{x}^{2}}-4}{-3}=\tan \dfrac{\pi }{4}$
By the standard value of trigonometric angle, we can write the value of $\tan \dfrac{\pi }{4}$ as 1.
So, $\dfrac{2{{x}^{2}}-4}{-3}=1$
Now on cross multiplication we get,
$2{{x}^{2}}-4=-3$
$\Rightarrow 2{{x}^{2}}=1$
$\Rightarrow {{x}^{2}}=\dfrac{1}{2}$
Hence, $x=\pm \dfrac{1}{\sqrt{2}}$
So, the value of x for which the equation satisfies is $\pm \dfrac{1}{\sqrt{2}}$ .

Note: While using identity ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ students should be careful about calculations so, as to avoid any mistakes. Any mistakes can lead to the whole solution being incorrect.