`cos^(-1)((3)/(5))+cos^(-1)((4)/(5))=`

Options

A. `(3pi)/(2)`
B. `(pi)/(2)`
C. `(-pi)/(2)`
D. `0`

Answer

To solve the problem \( \cos^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) \), we can use the properties of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Let \( \theta_1 = \cos^{-1}\left(\frac{3}{5}\right) \) and \( \theta_2 = \cos^{-1}\left(\frac{4}{5}\right) \)**:
\[ \theta_1 + \theta_2 = \cos^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) \]

2. **Use the identity for cosine of the sum of angles**:
We know that:
\[ \cos(\theta_1 + \theta_2) = \cos(\theta_1)\cos(\theta_2) - \sin(\theta_1)\sin(\theta_2) \]
Here, \( \cos(\theta_1) = \frac{3}{5} \) and \( \cos(\theta_2) = \frac{4}{5} \).

3. **Calculate \( \sin(\theta_1) \) and \( \sin(\theta_2) \)**:
Using the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \):
- For \( \theta_1 \):
\[ \sin(\theta_1) = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
- For \( \theta_2 \):
\[ \sin(\theta_2) = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

4. **Substitute the values into the cosine sum formula**:
\[ \cos(\theta_1 + \theta_2) = \left(\frac{3}{5}\right)\left(\frac{4}{5}\right) - \left(\frac{4}{5}\right)\left(\frac{3}{5}\right) \]
Simplifying this gives:
\[ \cos(\theta_1 + \theta_2) = \frac{12}{25} - \frac{12}{25} = 0 \]

5. **Determine the angle whose cosine is 0**:
Since \( \cos(\theta_1 + \theta_2) = 0 \), we have:
\[ \theta_1 + \theta_2 = \frac{\pi}{2} \]

### Final Answer:
\[ \cos^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2} \]