Step 1: Understanding the Concept:
To add different inverse trigonometric functions, it is best to convert them to a single common function, usually \( \tan^{-1} \), using the properties of right-angled triangles.
Step 2: Detailed Explanation:
1. Let \( \alpha = \sin^{-1} \frac{1}{\sqrt{5}} \). This means \( \sin \alpha = \frac{1}{\sqrt{5}} \).
Using Pythagoras: \( \text{Base} = \sqrt{(\sqrt{5})^{2} - 1^{2}} = \sqrt{4} = 2 \).
So, \( \tan \alpha = \frac{1}{2} \implies \alpha = \tan^{-1} \frac{1}{2} \).
2. Let \( \beta = \cos^{-1} \frac{3}{\sqrt{10}} \). This means \( \cos \beta = \frac{3}{\sqrt{10}} \).
Using Pythagoras: \( \text{Perp} = \sqrt{(\sqrt{10})^{2} - 3^{2}} = \sqrt{1} = 1 \).
So, \( \tan \beta = \frac{1}{3} \implies \beta = \tan^{-1} \frac{1}{3} \).
3. Now, add \( \alpha + \beta \):
\[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{1/2 + 1/3}{1 - (1/2 \times 1/3)} \right) \]
\[ = \tan^{-1} \left( \frac{5/6}{1 - 1/6} \right) = \tan^{-1} \left( \frac{5/6}{5/6} \right) = \tan^{-1} 1 = \frac{\pi}{4} \].
Step 3: Final Answer:
The value is \( \pi/4 \).