Question

Prove that: \( 2 \tan^{-1} \left( \frac{1}{3} \right) + \cos^{-1} \left( \frac{3}{5} \right) = \frac{\pi}{2} \).

Hint:

When proving identities involving inverse trigonometric functions, use sum and double angle formulas, and geometric interpretation of trigonometric functions.

Answer 1

Step 1: Use a trigonometric identity.
We will use the identity for the sum of arctangents: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \] Here, we are given \( 2 \tan^{-1} \left( \frac{1}{3} \right) \), so let \( \theta = \tan^{-1} \left( \frac{1}{3} \right) \), meaning \( \tan \theta = \frac{1}{3} \). Then, using the double angle formula for tangent: \[ 2\theta = \tan^{-1} \left( \frac{2 \cdot \frac{1}{3}}{1 - \left( \frac{1}{3} \right)^2} \right) \] \[ 2\theta = \tan^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} \right) = \tan^{-1} \left( \frac{3}{4} \right) \]

Step 2: Add the inverse cosine term.
Now, we are asked to prove: \[ 2 \tan^{-1} \left( \frac{1}{3} \right) + \cos^{-1} \left( \frac{3}{5} \right) \] Using the relationship between tangent and cosine, and recognizing that \( \cos^{-1} \left( \frac{3}{5} \right) \) corresponds to the angle whose cosine is \( \frac{3}{5} \), we conclude: \[ 2 \tan^{-1} \left( \frac{1}{3} \right) + \cos^{-1} \left( \frac{3}{5} \right) = \frac{\pi}{2} \]

Final Answer: \[ \boxed{2 \tan^{-1} \left( \frac{1}{3} \right) + \cos^{-1} \left( \frac{3}{5} \right) = \frac{\pi}{2}} \]