Question

Prove the result of the following expression, $4{{\tan }^{-1}}\dfrac{1}{5}-{{\tan }^{-1}}\dfrac{1}{239}=\dfrac{\pi }{4}$.

Answer 1

Hint:In order to prove the given equality, we should have some knowledge of a few inverse trigonometric formulas like, $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$. We also have to remember that ${{\tan }^{-1}}1=\dfrac{\pi }{4}$. We can prove the given equality by using these formulas.

Complete step-by-step answer:
In this question, we have been asked to prove the equality, $4{{\tan }^{-1}}\dfrac{1}{5}-{{\tan }^{-1}}\dfrac{1}{239}=\dfrac{\pi }{4}$. To prove this equality, we will first consider the left hand side or the LHS of the equation. So, we can write it as,
$LHS=4{{\tan }^{-1}}\dfrac{1}{5}-{{\tan }^{-1}}\dfrac{1}{239}$
Which can also be written as,
$LHS=2\left( 2{{\tan }^{-1}}\dfrac{1}{5} \right)-{{\tan }^{-1}}\dfrac{1}{239}$
Now, we know that $2{{\tan }^{-1}}x$ can be expressed as ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. So, by using this formula, we get the LHS as,
$LHS=2\left[ {{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right) \right]-{{\tan }^{-1}}\dfrac{1}{239}$
Now, we know that, ${{\left( \dfrac{1}{5} \right)}^{2}}=\dfrac{1}{25}$. So, by applying it, we will simplify the LHS as follows,
$\begin{align}
  & LHS=2{{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{1-\dfrac{1}{25}} \right)-{{\tan }^{-1}}\dfrac{1}{239} \\
 & \Rightarrow LHS=2{{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{\dfrac{25-1}{25}} \right)-{{\tan }^{-1}}\dfrac{1}{239} \\
 & \Rightarrow LHS=2{{\tan }^{-1}}\left( \dfrac{2\times 25}{24\times 5} \right)-{{\tan }^{-1}}\dfrac{1}{239} \\
\end{align}$
We can further write it as,
$LHS=2{{\tan }^{-1}}\left( \dfrac{5}{12} \right)-{{\tan }^{-1}}\dfrac{1}{239}$
Now, we will again use the formula, $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$, to simplify the above expression further. So, applying this, we can write the LHS as,
$LHS={{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{5}{12} \right)}{1-{{\left( \dfrac{5}{12} \right)}^{2}}} \right)-{{\tan }^{-1}}\dfrac{1}{239}$
Now, we will simplify it further by writing, ${{\left( \dfrac{5}{12} \right)}^{2}}=\dfrac{25}{144}$. So, we get the LHS as,
$\begin{align}
  & LHS={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{5}{6} \right)}{1-\left( \dfrac{25}{144} \right)} \right)-{{\tan }^{-1}}\dfrac{1}{239} \\
 & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{5}{6}}{\dfrac{144-25}{144}} \right]-{{\tan }^{-1}}\dfrac{1}{239} \\
 & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{5\times 144}{6\times 119} \right]-{{\tan }^{-1}}\dfrac{1}{239} \\
\end{align}$
And we can further write it as,
$LHS={{\tan }^{-1}}\left[ \dfrac{120}{119} \right]-{{\tan }^{-1}}\dfrac{1}{239}$
Now, we know that ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$, so, we can apply this formula and write the above expression as,
$LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{120}{119}-\dfrac{1}{239}}{1+\dfrac{120}{119}\times \dfrac{1}{239}} \right]$
Now, we will simplify it by taking the LCM of the numerator and the denominator. Hence, we can write the LHS as,
$\begin{align}
  & LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{120\times 239-119}{119\times 239}}{\dfrac{119\times 239+120}{119\times 239}} \right] \\
 & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{\left( 120\times 239-119 \right)\left( 119\times 239 \right)}{\left( 119\times 239+120 \right)\left( 119\times 239 \right)} \right] \\
 & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{\left( 120\times 239-119 \right)}{\left( 119\times 239+120 \right)} \right] \\
 & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{28561}{28561} \right] \\
\end{align}$
We know that the common terms of the numerator and the denominator gets cancelled, so we get the LHS as,
$LHS={{\tan }^{-1}}1$
And from the trigonometric ratios, we know that $\tan \dfrac{\pi }{4}=1$, which can be expressed as, $\dfrac{\pi }{4}={{\tan }^{-1}}1$. Hence, we can write, $LHS=\dfrac{\pi }{4}$, which is the same as the right hand side or the RHS of the expression given in the question, so LHS = RHS.
Hence, we have proved the expression given in the question.

Note: There is a possibility of making calculation mistakes while solving this question as it involves a lot of calculations. And we, know that we have to get ${{\tan }^{-1}}1=\dfrac{\pi }{4}$ in the end, so we will come to know if we are on the right track in the solving of the question. Also, we should remember the formulas like, $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$ to solve the question.