Question

If $\tan^{-1} x + \tan^{-1} y = \frac{2\pi}{3}$, then the value of $\cot^{-1} x + \cot^{-1} y$ is equal to:

• A. $\frac{\pi}{2}$
• B. $\frac{1}{2}$
• C. $\frac{\pi}{3}$
• D. $\frac{\sqrt{3}}{2}$
• E. $\pi$

Hint:

This technique is called the "Complementary Angle Method." It works for any pair of inverse functions $(\sin^{-1} + \cos^{-1})$ or $(\sec^{-1} + \csc^{-1})$. The sum of the "co-" functions and their originals is always $\pi$.

Answer 1

The Correct Option is C

Concept: Inverse trigonometric functions have complementary properties. Specifically, for any real value $z$, the sum of its inverse tangent and inverse cotangent is a constant $\pi/2$. We can leverage this relationship to transform an equation in $\tan^{-1}$ into one in $\cot^{-1}$ without needing to solve for $x$ or $y$ individually.

Step 1: State the fundamental identity for each variable.
For the variable $x$: $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \Rightarrow \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x$
For the variable $y$: $\tan^{-1} y + \cot^{-1} y = \frac{\pi}{2} \Rightarrow \cot^{-1} y = \frac{\pi}{2} - \tan^{-1} y$

Step 2: Formulate the sum $\cot^{-1} x + \cot^{-1} y$.
Add the two expressions derived in
Step 1: \[ \cot^{-1} x + \cot^{-1} y = \left( \frac{\pi}{2} - \tan^{-1} x \right) + \left( \frac{\pi}{2} - \tan^{-1} y \right) \] Combine the constant terms: \[ \cot^{-1} x + \cot^{-1} y = \pi - (\tan^{-1} x + \tan^{-1} y) \]

Step 3: Substitute the given numerical sum.
The problem states that $\tan^{-1} x + \tan^{-1} y = \frac{2\pi}{3}$. Substituting this into our formula: \[ \cot^{-1} x + \cot^{-1} y = \pi - \frac{2\pi}{3} \] To perform the subtraction, use a common denominator of 3: \[ \frac{3\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{3} \]