`tan^(-1)((1+sqrt3)/(3+sqrt3))+sec^-1(sqrt((8+4sqrt3)/(6+3sqrt3)))` is equal to

Options

A. `pi/2`
B. `pi/4`
C. `pi/6`
D. `pi/3`

Answer

To solve the expression \( \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right) \), we will break it down into manageable steps.

### Step 1: Simplify the first term \( \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) \)

We start with the expression inside the \( \tan^{-1} \):

\[ \frac{1+\sqrt{3}}{3+\sqrt{3}} \]

To simplify this, we can multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{(1+\sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} \]

Calculating the denominator:

\[ (3+\sqrt{3})(3-\sqrt{3}) = 9 - 3 = 6 \]

Calculating the numerator:

\[ (1+\sqrt{3})(3-\sqrt{3}) = 3 - \sqrt{3} + 3\sqrt{3} - 3 = 2\sqrt{3} \]

Thus, we have:

\[ \frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \]

Now we can rewrite the first term:

\[ \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \]

### Step 2: Identify the angle

We know that:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3} \]

Thus:

\[ \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \]

### Step 3: Simplify the second term \( \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right) \)

We simplify the expression inside the \( \sec^{-1} \):

\[ \sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}} \]

Calculating the numerator and denominator separately:

1. **Numerator**: \( 8 + 4\sqrt{3} \)
2. **Denominator**: \( 6 + 3\sqrt{3} \)

To simplify, we can multiply both the numerator and denominator by the conjugate of the denominator:

\[ \frac{(8+4\sqrt{3})(6-3\sqrt{3})}{(6+3\sqrt{3})(6-3\sqrt{3})} \]

Calculating the denominator:

\[ (6+3\sqrt{3})(6-3\sqrt{3}) = 36 - 27 = 9 \]

Calculating the numerator:

\[ (8+4\sqrt{3})(6-3\sqrt{3}) = 48 - 24\sqrt{3} + 24\sqrt{3} - 12 = 36 \]

Thus, we have:

\[ \sqrt{\frac{36}{9}} = \sqrt{4} = 2 \]

### Step 4: Rewrite the second term

Now we can rewrite the second term:

\[ \sec^{-1}(2) \]

### Step 5: Identify the angle

We know that:

\[ \sec\left(\frac{\pi}{3}\right) = 2 \]

Thus:

\[ \sec^{-1}(2) = \frac{\pi}{3} \]

### Step 6: Combine both terms

Now we can combine both terms:

\[ \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right) = \frac{\pi}{6} + \frac{\pi}{3} \]

### Step 7: Find a common denominator

To add these fractions, we convert \( \frac{\pi}{3} \) to sixths:

\[ \frac{\pi}{3} = \frac{2\pi}{6} \]

Thus:

\[ \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]

### Final Answer

The final answer is:

\[ \frac{\pi}{2} \]