Question

Solve the following equation, ${{\tan }^{-1}}\dfrac{1}{4}+2{{\tan }^{-1}}\dfrac{1}{5}+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$.

Answer 1

Hint:To prove the equation given in the question, we should have some knowledge of a few inverse trigonometric formulas like, ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ and $2{{\tan }^{-1}}a={{\tan }^{-1}}\left( \dfrac{2a}{1-{{a}^{2}}} \right)$. We also should remember a few standard trigonometric angles like $\tan \dfrac{\pi }{4}=1$. We can prove the given equation by using these formulas.

Complete step-by-step answer:
In this question, we have been asked to solve the value of x in the equation, ${{\tan }^{-1}}\dfrac{1}{4}+2{{\tan }^{-1}}\dfrac{1}{5}+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$. Now, we know that $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. So, we can write the given equality for $x=\dfrac{1}{5}$ as,
${{\tan }^{-1}}\dfrac{1}{4}+{{\tan }^{-1}}\left[ \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$
Now, we will simplify it further to get,
$\begin{aligned} & {{\tan }^{-1}}\dfrac{1}{4}+{{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{2}{5} \right)}{1-\left( \dfrac{1}{25} \right)} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\dfrac{1}{4}+{{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{2}{5} \right)}{\dfrac{25-1}{25}} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\dfrac{1}{4}+{{\tan }^{-1}}\left[ \dfrac{2\times 25}{5\times 24} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\dfrac{1}{4}+{{\tan }^{-1}}\left[ \dfrac{5}{12} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ \end{aligned}$
Now, we know that ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. So, for $a=\dfrac{1}{4}$ and $b=\dfrac{5}{12}$, we get the equation as,
${{\tan }^{-1}}\left[ \dfrac{\dfrac{1}{4}+\dfrac{5}{12}}{1-\dfrac{1}{4}\times \dfrac{5}{12}} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$
And we can further simplify it as,
$\begin{aligned} & {{\tan }^{-1}}\left[ \dfrac{\dfrac{3+5}{12}}{\dfrac{48-5}{48}} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{8\times 48}{12\times 43} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{32}{43} \right]+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ \end{aligned}$
We will again apply the formula ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. So, for $a=\dfrac{32}{43}$ and $b=\dfrac{1}{6}$, we get the equation as,
${{\tan }^{-1}}\left[ \dfrac{\dfrac{32}{43}+\dfrac{1}{6}}{1-\dfrac{32}{43}\times \dfrac{1}{6}} \right]+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$
And we can simplify it further as,
$\begin{aligned} & {{\tan }^{-1}}\left[ \dfrac{\dfrac{192+43}{43\times 6}}{\dfrac{258-32}{43\times 6}} \right]+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{235\times 258}{226\times 258} \right]+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{235}{226} \right]+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4} \\\ \end{aligned}$
We know that $\tan \dfrac{\pi }{4}=1$. So, we can write $\dfrac{\pi }{4}={{\tan }^{-1}}1$. Therefore, we get the equation as,
${{\tan }^{-1}}\left[ \dfrac{235}{226} \right]+{{\tan }^{-1}}\dfrac{1}{x}={{\tan }^{-1}}1$
And if we keep the terms with x on one side and the others on the other side, then we get the above equation as follows,
${{\tan }^{-1}}\dfrac{1}{x}={{\tan }^{-1}}1-{{\tan }^{-1}}\left[ \dfrac{235}{226} \right]$
Now, we know that ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$. So, applying that formula for $a=1$ and $b=\dfrac{235}{226}$, we get the equation as,
${{\tan }^{-1}}\dfrac{1}{x}={{\tan }^{-1}}\left[ \dfrac{1-\dfrac{235}{226}}{1+1\times \dfrac{235}{226}} \right]$
And on further simplification, we get,
$\begin{aligned} & {{\tan }^{-1}}\dfrac{1}{x}={{\tan }^{-1}}\left[ \dfrac{\dfrac{226-235}{226}}{\dfrac{226+235}{226}} \right] \\\ & \Rightarrow {{\tan }^{-1}}\dfrac{1}{x}={{\tan }^{-1}}\left[ \dfrac{-9\times 226}{461\times 226} \right] \\\ \end{aligned}$
Now, we will take the tangent ratio of the equality. So, we get,
$\tan \left( {{\tan }^{-1}}\dfrac{1}{x} \right)=\tan \left( {{\tan }^{-1}}\left( \dfrac{-9\times 226}{461\times 226} \right) \right)$
And we know that $\tan \left( {{\tan }^{-1}}\alpha \right)=\alpha$. So, we get the equation as,
$\begin{aligned} & \dfrac{1}{x}=\dfrac{-9}{461} \\\ & \Rightarrow x=\dfrac{-461}{9} \\\ \end{aligned}$
Hence, we can say that we get $x=\dfrac{-461}{9}$ for the equation, ${{\tan }^{-1}}\dfrac{1}{4}+2{{\tan }^{-1}}\dfrac{1}{5}+{{\tan }^{-1}}\dfrac{1}{6}+{{\tan }^{-1}}\dfrac{1}{x}=\dfrac{\pi }{4}$.

Note: There is a possibility of calculation mistakes in this question due to the long calculations involved. Also, sometimes, we write the wrong formula of ${{\tan }^{-1}}a+{{\tan }^{-1}}b$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b$, which are, ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$. So, one should be careful in writing the formulas.