Question

Evaluate: \(\tan^{-1}(1) + \tan^{-1}(4) + \tan^{-1}(5) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\). Find the value of \(\alpha\).

Hint:

When combining several $\tan^{-1}$ terms, add or subtract them two at a time using the standard formulas. Before accepting a sum, check whether the product of the two arguments exceeds 1 — if it does, you must add $\pi$ to stay in the correct range. Also watch for values that are reciprocals of each other, since their sum simplifies immediately to $\pi/2$.

Answer 1

Step 1: Understanding the Concept:
We use the property that \(\tan^{-1}(x) + \tan^{-1}(1/x) = \pi/2\) for \(x>0\). This simplifies the series significantly.

Step 2: Key Formula or Approach:
1. \(\tan^{-1}(1) = \pi/4\).
2. \(\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})\) (used here for subtraction).

Step 3: Detailed Explanation:
Let the LHS be \(L\):
\[ L = \tan^{-1}(1) + \tan^{-1}(4) + \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}(5) \] Using \(\tan^{-1}(4) + \tan^{-1}(1/4) = \pi/2\):
\[ L = \frac{\pi}{4} + \frac{\pi}{2} + \tan^{-1}(5) = \frac{3\pi}{4} + \tan^{-1}(5) \] The equation is \(L = \pi + \tan^{-1}(\alpha/2)\).
\[ \frac{3\pi}{4} + \tan^{-1}(5) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right) \] \[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}(5) - \frac{\pi}{4} \] Since \(\pi/4 = \tan^{-1}(1)\):
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}(5) - \tan^{-1}(1) \] Using the formula \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}(\frac{x-y}{1+xy})\):
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}\left(\frac{5 - 1}{1 + 5(1)}\right) = \tan^{-1}\left(\frac{4}{6}\right) = \tan^{-1}\left(\frac{2}{3}\right) \] By comparison:
\[ \frac{\alpha}{2} = \frac{2}{3} \implies \alpha = \frac{4}{3} \]
Step 4: Final Answer:
The value of \(\alpha\) is \(\frac{4}{3}\).

Answer 2

Concept:

• $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$ when $x,y>0$; if additionally $xy>1$, add $\pi$ to the result, since the raw value falls outside the principal range of $\tan^{-1}$.
• $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\dfrac{x-y}{1+xy}\right)$ for $x,y>0$, with no correction needed.
• Terms can be combined two at a time in any convenient order, as long as the branch correction rule is applied whenever it is needed.

Step 1: Combine $\tan^{-1}(1)$ and $\tan^{-1}(4)$ first.
Here $xy = 1 \times 4 = 4 > 1$, so the correction applies:
$\tan^{-1}(1) + \tan^{-1}(4) = \pi + \tan^{-1}\left(\dfrac{1+4}{1-4}\right) = \pi + \tan^{-1}\left(\dfrac{5}{-3}\right) = \pi - \tan^{-1}\left(\dfrac{5}{3}\right)$.

Step 2: Add $\tan^{-1}(5)$ to this result.
$\pi - \tan^{-1}\left(\dfrac{5}{3}\right) + \tan^{-1}(5) = \pi + \left[\tan^{-1}(5) - \tan^{-1}\left(\dfrac{5}{3}\right)\right]$.
Since $5 > 5/3 > 0$, no correction is needed:
$\tan^{-1}(5) - \tan^{-1}\left(\dfrac{5}{3}\right) = \tan^{-1}\left(\dfrac{5 - 5/3}{1 + 5 \cdot 5/3}\right) = \tan^{-1}\left(\dfrac{10/3}{28/3}\right) = \tan^{-1}\left(\dfrac{5}{14}\right)$.
So the running total is $\pi + \tan^{-1}\left(\dfrac{5}{14}\right)$.

Step 3: Add the last term, $\tan^{-1}(1/4)$.
Here $xy = \dfrac{5}{14} \times \dfrac{1}{4} = \dfrac{5}{56} < 1$, so no correction is needed:
$\tan^{-1}\left(\dfrac{5}{14}\right) + \tan^{-1}\left(\dfrac{1}{4}\right) = \tan^{-1}\left(\dfrac{5/14 + 1/4}{1 - 5/56}\right) = \tan^{-1}\left(\dfrac{17/28}{51/56}\right) = \tan^{-1}\left(\dfrac{2}{3}\right)$.
Total: $L = \pi + \tan^{-1}\left(\dfrac{2}{3}\right)$.

Step 4: Compare with the given form.
$\pi + \tan^{-1}\left(\dfrac{\alpha}{2}\right) = \pi + \tan^{-1}\left(\dfrac{2}{3}\right) \Rightarrow \dfrac{\alpha}{2} = \dfrac{2}{3} \Rightarrow \alpha = \dfrac{4}{3}$.

Final Answer: $\alpha = \dfrac{4}{3}$.