Question

If \(\tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\), find \(\alpha\).

Hint:

The condition \(xy>1\) is the most common trap in MHT-CET inverse trig questions. If you forget to add \(\pi\), your answer will be off by exactly \(\pi\) (roughly 3.14), which is usually one of the wrong options.

Answer 1

Step 1: Understanding the Concept:
This involves inverse trigonometric identities. Specifically, we must be careful with \(\tan^{-1}x + \tan^{-1}y\) when \(xy>1\), as it introduces a \(\pi\) term.

Step 2: Key Formula or Approach:
1. \(\tan^{-1}(-x) = -\tan^{-1}x\).
2. If \(xy>1\), then \(\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\).
3. If \(xy<1\), then \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\).

Step 3: Detailed Explanation:
LHS \(= \tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}(1/4)\).
We know \(\tan^{-1}(-1) = -\frac{\pi}{4}\).
Now evaluate \(\tan^{-1}(5) + \tan^{-1}(3)\). Here \(5 \times 3 = 15>1\).
\[ \tan^{-1}(5) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{5+3}{1-15}\right) = \pi + \tan^{-1}\left(-\frac{8}{14}\right) = \pi - \tan^{-1}\left(\frac{4}{7}\right) \] Substitute back into LHS:
\[ LHS = -\frac{\pi}{4} + \pi - \tan^{-1}\left(\frac{4}{7}\right) + \tan^{-1}\left(\frac{1}{4}\right) = \frac{3\pi}{4} + \tan^{-1}\left(\frac{1/4 - 4/7}{1 + (1/4)(4/7)}\right) \] \[ LHS = \frac{3\pi}{4} + \tan^{-1}\left(\frac{(7-16)/28}{1 + 1/7}\right) = \frac{3\pi}{4} + \tan^{-1}\left(\frac{-9/28}{8/7}\right) = \frac{3\pi}{4} - \tan^{-1}\left(\frac{9}{32}\right) \] We are given: \(\frac{3\pi}{4} - \tan^{-1}\left(\frac{9}{32}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\).
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = -\frac{\pi}{4} - \tan^{-1}\left(\frac{9}{32}\right) = - \left[ \tan^{-1}(1) + \tan^{-1}\left(\frac{9}{32}\right) \right] \] Since \(1 \times (9/32)<1\):
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = -\tan^{-1}\left(\frac{1 + 9/32}{1 - 9/32}\right) = -\tan^{-1}\left(\frac{41/32}{23/32}\right) = \tan^{-1}\left(-\frac{41}{23}\right) \] Comparing: \(\frac{\alpha}{2} = -\frac{41}{23} \implies \alpha = -\frac{82}{23}\).

Step 4: Final Answer:
The value of \(\alpha\) is \(-\frac{82}{23}\).