The value of expression `tan^(- 1)((sqrt(2))/2)+sin^(- 1)((sqrt(5))/5)-cos^(- 1)((sqrt(10))/10)`

Options

A. `cot^(-1) ((1+sqrt2)/(1-sqrt2))`
B. `cot^(-1) ((sqrt2 + 1)/(sqrt2-1))`
C. `- pi + cot^(-1) ((1+sqrt2)/(1 - sqrt2))`
D. `pi - cot^(-1) ((1-sqrt2)/(1 + sqrt2))`

Answer

To solve the expression \( \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) + \sin^{-1}\left(\frac{\sqrt{5}}{5}\right) - \cos^{-1}\left(\frac{\sqrt{10}}{10}\right) \), we will break it down step by step.

### Step 1: Simplify the Inverse Trigonometric Functions
We start by rewriting the inverse trigonometric functions in terms of their equivalent angles.

1. **For \( \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) \)**:
\[ \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \]

2. **For \( \sin^{-1}\left(\frac{\sqrt{5}}{5}\right) \)**:
Let \( y = \sin^{-1}\left(\frac{\sqrt{5}}{5}\right) \). Therefore, \( \sin y = \frac{\sqrt{5}}{5} \).
- In a right triangle, if the opposite side is \( \sqrt{5} \) and the hypotenuse is \( 5 \), the adjacent side can be calculated using the Pythagorean theorem:
\[ \text{Adjacent} = \sqrt{5^2 - (\sqrt{5})^2} = \sqrt{25 - 5} = \sqrt{20} = 2\sqrt{5} \]
- Therefore, \( \tan y = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{5}}{2\sqrt{5}} = \frac{1}{2} \).
- Hence, \( y = \tan^{-1}\left(\frac{1}{2}\right) \).

3. **For \( \cos^{-1}\left(\frac{\sqrt{10}}{10}\right) \)**:
Let \( z = \cos^{-1}\left(\frac{\sqrt{10}}{10}\right) \). Therefore, \( \cos z = \frac{\sqrt{10}}{10} \).
- In a right triangle, if the adjacent side is \( \sqrt{10} \) and the hypotenuse is \( 10 \), the opposite side can be calculated:
\[ \text{Opposite} = \sqrt{10^2 - (\sqrt{10})^2} = \sqrt{100 - 10} = \sqrt{90} = 3\sqrt{10} \]
- Therefore, \( \tan z = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3\sqrt{10}}{\sqrt{10}} = 3 \).
- Hence, \( z = \tan^{-1}(3) \).

### Step 2: Substitute Back into the Expression
Now substituting back into the expression:
\[ \frac{\pi}{4} + \tan^{-1}\left(\frac{1}{2}\right) - \tan^{-1}(3) \]

### Step 3: Use the Formula for \( \tan^{-1} \)
Using the formula for the difference of two inverse tangents:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{if } ab < 1 \]
We can rewrite:
\[ \tan^{-1}\left(\frac{1}{2}\right) - \tan^{-1}(3) = \tan^{-1}\left(\frac{\frac{1}{2} - 3}{1 + \frac{1}{2} \cdot 3}\right) \]
Calculating:
- \( a = \frac{1}{2} \)
- \( b = 3 \)

Calculating the numerator:
\[ \frac{1}{2} - 3 = \frac{1 - 6}{2} = \frac{-5}{2} \]

Calculating the denominator:
\[ 1 + \frac{3}{2} = \frac{2 + 3}{2} = \frac{5}{2} \]

Thus:
\[ \tan^{-1}\left(\frac{-5/2}{5/2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Step 4: Combine the Results
Now we have:
\[ \frac{\pi}{4} - \frac{\pi}{4} = 0 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{0} \]