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 Question:
This is a simplification problem for inverse trigonometric functions.
Step 2: Key Formula or Approach:
1. \( \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) if \( xy>1, x,y>0 \).
2. \( \tan^{-1}(-x) = -\tan^{-1}x \).
Step 3: Detailed Explanation:
LHS \( = \tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}(1/4) \)
\( = -\frac{\pi}{4} + [\pi + \tan^{-1}(\frac{5+3}{1-15})] + \tan^{-1}(1/4) \)
\( = -\frac{\pi}{4} + \pi + \tan^{-1}(-8/14) + \tan^{-1}(1/4) \)
\( = \frac{3\pi}{4} - \tan^{-1}(4/7) + \tan^{-1}(1/4) \)
\( = \pi - \tan^{-1}(1) - \tan^{-1}(4/7) + \tan^{-1}(1/4) \)
Combine terms:
\( = \pi - [\tan^{-1}(1) + \tan^{-1}(4/7)] + \tan^{-1}(1/4) \)
\( = \pi - \tan^{-1}(\frac{1+4/7}{1-4/7}) + \tan^{-1}(1/4) \)
\( = \pi - \tan^{-1}(11/3) + \tan^{-1}(1/4) \)
\( = \pi + \tan^{-1}(\frac{1/4 - 11/3}{1 + (1/4)(11/3)}) \)
\( = \pi + \tan^{-1}(\frac{(3-44)/12}{(12+11)/12}) = \pi + \tan^{-1}(-41/23) \)
Comparing with \( \pi + \tan^{-1}(\alpha/2) \):
\( \alpha/2 = -41/23 \Rightarrow \alpha = -82/23 \).
Step 4: Final Answer:
The value of \( \alpha \) is \( -82/23 \).