Question

$\cos^{-1}\left(\frac{4}{5}\right)+\cos^{-1}\left(\frac{12}{13}\right)=$

• A. $\cos^{-1}\left(\frac{24}{25}\right)$
• B. $\cos^{-1}\left(\frac{33}{65}\right)$
• C. $\cos^{-1}\left(\frac{5}{13}\right)$
• D. $\cos^{-1}\left(\frac{3}{5}\right)$

Hint:

You can also solve this by converting the inverse cosines to inverse tangents using Pythagorean triplets: $\cos^{-1}(4/5) = \tan^{-1}(3/4)$ and $\cos^{-1}(12/13) = \tan^{-1}(5/12)$. Then apply the $\tan^{-1}A + \tan^{-1}B$ formula!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the sum of two inverse cosine functions.

Step 2: Key Formula or Approach:
The standard addition formula for inverse cosine functions is:
$$\cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right)$$ for $x, y > 0$.

Step 3: Detailed Explanation:
Let $x = \frac{4}{5}$ and $y = \frac{12}{13}$.
Calculate the individual components for the formula:
$$xy = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) = \frac{48}{65}$$ $$\sqrt{1-x^2} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$ $$\sqrt{1-y^2} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$$ Now, substitute these into the addition formula:
$$\cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \cos^{-1}\left( \frac{48}{65} - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \right)$$ $$= \cos^{-1}\left( \frac{48}{65} - \frac{15}{65} \right)$$ $$= \cos^{-1}\left( \frac{33}{65} \right)$$

Step 4: Final Answer:
The evaluated expression is $\cos^{-1}\left(\frac{33}{65}\right)$, matching option (B).