Q.Find the value of $4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}$.

Concept: Inverse Tangent Identity — we use the formula

$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$ (valid when $xy<1$).

Step 1: Apply the formula to $2\tan^{-1}\frac{1}{5}$:

Step 2: Now $4\tan^{-1}\frac{1}{5} = 2\left(2\tan^{-1}\frac{1}{5}\right) = 2\tan^{-1}\frac{5}{12}$. Apply the formula again:

Step 3: Subtract $\tan^{-1}\frac{1}{239}$:

$\tan^{-1}\frac{120}{119} - \tan^{-1}\frac{1}{239} = \tan^{-1}\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119}\cdot\frac{1}{239}}.$ …

This is a classic problem — it’s the heart of John Machin’s 1706 formula for $\pi$, which he used to compute $\pi$ to 100 decimal places. The trick is to notice that $4\tan^{-1}\frac{1}{5}$ is not just four copies of the same angle; it’s an invitation to use the tangent addition formula twice.

1. The core tool: adding inverse tangents

For any two numbers $a$ and $b$ (with $ab \neq 1$ and the sum of the angles in the correct quadrant), we have:

Why? Because if $\alpha = \tan^{-1}a$ and $\beta = \tan^{-1}b$, then $\tan(\alpha+\beta) = \frac{a+b}{1-ab}$. The inverse tangent then recovers the angle, provided we stay within $(-\pi/2, \pi/2)$ — which we will, since all angles here are small.

2. First double: $2\tan^{-1}\frac{1}{5}$

Let $a = b = \frac{1}{5}$. Then:

So $2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{5}{12}$.

3. Second double: $4\tan^{-1}\frac{1}{5}$

Now add another $\tan^{-1}\frac{1}{5}$ to the result. But it’s easier to double again: take $a = b = \frac{5}{12}$.

Thus:

4. Subtract $\tan^{-1}\frac{1}{239}$

We now have: …

Method: Collapsing multiples of inverse tangent by repeated doubling

To evaluate an expression like $4\tan^{-1}a-\tan^{-1}b$, reduce the large multiple step by step with the doubling identity, then combine with the subtraction formula.

Steps

Step 1: Double once.

Step 2: Double the result to reach $4\tan^{-1}a$.

Feed the value from Step 1 back through the same formula: $4\tan^{-1}a=2\big(2\tan^{-1}a\big)$.

Step 3: Subtract the remaining term.

$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\!\left(\frac{A-B}{1+AB}\right).$ …

Common Mistakes

Mistake 1: Treating $4\tan^{-1}\tfrac15$ as $\tan^{-1}\tfrac45$.

Why it's wrong: the $4$ multiplies the angle, not the argument. Correct approach: apply the doubling identity twice — $2\tan^{-1}\tfrac15=\tan^{-1}\tfrac{5}{12}$, then $2\tan^{-1}\tfrac{5}{12}=\tan^{-1}\tfrac{120}{119}$.

Mistake 2: Using the addition (sign $-$) formula for the subtraction, or vice-versa.

Why it's wrong: $\tan^{-1}A-\tan^{-1}B$ needs $\tfrac{A-B}{1+AB}$, with $+AB$ in the denominator. Correct approach: with $A=\tfrac{120}{119}$, $B=\tfrac{1}{239}$, numerator and denominator both equal $\tfrac{28561}{119\cdot239}$, so the ratio is $1$. …