Question

Find the value of x, if:
${{\tan }^{-1}}\dfrac{x-2}{x-1}+{{\tan }^{-1}}\dfrac{x+2}{x+1}=\dfrac{\pi }{4}.$

Answer 1

Hint: We will use the formula for ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ , which is given by ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$. The terms in the LHS of the question can be expanded using this formula. We have to substitute $A=\dfrac{x-2}{x-1}\ and\ B=\dfrac{x+2}{x+1}$ in the above formula to expand the terms given in the question. After that, we will equate the result to the RHS, i.e. $\dfrac{\pi }{4}$. Then we can take tan on both sides. After doing this, we will end up with a quadratic equation in x which can be solved to get the value of x.

Complete step-by-step solution -
Given that;
${{\tan }^{-1}}\dfrac{x-2}{x-1}+{{\tan }^{-1}}\dfrac{x+2}{x+1}=\dfrac{\pi }{4}.$…………………. (1)
And, we have to find the value of x so that it will satisfy this equation.
We know that;
${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$
Applying this formula in our question, we get;
$\begin{aligned} & {{\tan }^{-1}}\left( \dfrac{x-2}{x-1} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+1} \right)={{\tan }^{-1}}\dfrac{\left( \left( \dfrac{x-2}{x-1} \right)+\left( \dfrac{x+2}{x+1} \right) \right)}{\left( 1-\left( \dfrac{x-2}{x-1} \right).\left( \dfrac{x+2}{x+1} \right) \right)} \\\ & {{\tan }^{-1}}\left( \dfrac{x-2}{x-1} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+1} \right)={{\tan }^{-1}}\dfrac{\dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)}}{1-\dfrac{{{x}^{2}}-{{\left( 2 \right)}^{2}}}{{{\left( x \right)}^{2}}-1}} \\\ \end{aligned}$
As we know;

& {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{x-2}{x-1} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+1} \right)={{\tan }^{-1}}\dfrac{\dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{{{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}}}}{\dfrac{{{x}^{2}}-1-{{x}^{2}}+4}{{{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}}}} \\\ \end{aligned}$$ Cancelling the similar terms, we get; $${{\tan }^{-1}}\left( \dfrac{x-2}{x-1} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+1} \right)={{\tan }^{-1}}\left\\{ \dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{3} \right\\}$$ Putting the value of; $${{\tan }^{-1}}\left( \dfrac{x-2}{x-1} \right)+{{\tan }^{-1}}\left( \dfrac{x+2}{x+1} \right)={{\tan }^{-1}}\left\\{ \dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{3} \right\\}$$ in the equation (1), we get; $$={{\tan }^{-1}}\left\\{ \dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{3} \right\\}=\dfrac{\pi }{4}$$ Transposing $${{\tan }^{-1}}$$from LHS to RHS, we get; $$=\dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{3}=\tan \left( \dfrac{\pi }{4} \right)$$………….. (2) We know the value of $$\tan \dfrac{\pi }{4}=1$$. Putting the value of $$\tan \left( \dfrac{\pi }{4} \right)=1$$in equation (2), we get; $$\begin{aligned} & =\dfrac{\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)}{3}=1 \\\ & =\left( x-2 \right)\left( x+1 \right)+\left( x+2 \right)\left( x-1 \right)=3 \\\ & ={{x}^{2}}+x-2x-2+{{x}^{2}}-x+2x-2=3 \\\ & ={{x}^{2}}+{{x}^{2}}+x-x-2x+2x-4=3 \\\ \end{aligned}$$ Cancelling the similar terms, we get; $\begin{aligned} & =2{{x}^{2}}-4=3 \\\ & =2{{x}^{2}}=3+4 \\\ & =2{{x}^{2}}=7 \\\ & ={{x}^{2}}=\dfrac{7}{2} \\\ & =x=\pm \sqrt{\dfrac{7}{2}} \\\ \end{aligned}$ Note: Many time students confuse with the formula ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\ or\ {{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)$. Because of minus signs as a result they take wrong formula which leads to the formation of wrong answers. Therefore, it is recommended to know all the formulas to solve such problems quickly. It is also recommended to go through step by step inorder to avoid any silly mistakes.