`2tan^(-1) ""1/5 + sec^(-1) ""(5sqrt2)/7 +2 tan^(-1)"1/8=`

Options

A. `pi//2`
B. `-pi//2`
C. `pi//4`
D. `-pi//4`

Answer

To solve the expression \(2\tan^{-1}\left(\frac{1}{5}\right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) + 2\tan^{-1}\left(\frac{1}{8}\right)\), we can break it down step by step.

### Step 1: Combine the first and third terms
We have:
\[ 2\tan^{-1}\left(\frac{1}{5}\right) + 2\tan^{-1}\left(\frac{1}{8}\right) \]
We can factor out the 2:
\[ 2\left(\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right)\right) \]

### Step 2: Use the formula for the sum of arctangents
The formula for the sum of two arctangents is:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
where \(x = \frac{1}{5}\) and \(y = \frac{1}{8}\).

Calculating \(x + y\):
\[ x + y = \frac{1}{5} + \frac{1}{8} = \frac{8 + 5}{40} = \frac{13}{40} \]

Calculating \(1 - xy\):
\[ xy = \frac{1}{5} \cdot \frac{1}{8} = \frac{1}{40} \]
\[ 1 - xy = 1 - \frac{1}{40} = \frac{40 - 1}{40} = \frac{39}{40} \]

Now, substituting these values into the formula:
\[ \tan^{-1}\left(\frac{\frac{13}{40}}{\frac{39}{40}}\right) = \tan^{-1}\left(\frac{13}{39}\right) \]

### Step 3: Substitute back into the expression
Now, we have:
\[ 2\tan^{-1}\left(\frac{13}{39}\right) \]

### Step 4: Use the double angle formula for arctangent
The double angle formula for arctangent is:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
Let \(x = \frac{13}{39}\):
\[ 2\tan^{-1}\left(\frac{13}{39}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{13}{39}}{1 - \left(\frac{13}{39}\right)^2}\right) \]

Calculating \(2x\):
\[ 2x = \frac{26}{39} = \frac{2}{3} \]

Calculating \(1 - x^2\):
\[ x^2 = \left(\frac{13}{39}\right)^2 = \frac{169}{1521} \]
\[ 1 - x^2 = 1 - \frac{169}{1521} = \frac{1521 - 169}{1521} = \frac{1352}{1521} \]

Now we substitute these values:
\[ \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{1352}{1521}}\right) = \tan^{-1}\left(\frac{2 \cdot 1521}{3 \cdot 1352}\right) = \tan^{-1}\left(\frac{3042}{4056}\right) \]

### Step 5: Simplify the fraction
Now we simplify \(\frac{3042}{4056}\):
\[ \frac{3042 \div 2}{4056 \div 2} = \frac{1521}{2028} \]
Continuing to simplify:
\[ \frac{1521 \div 3}{2028 \div 3} = \frac{507}{676} \]

### Step 6: Add the second term
Now, we need to add the second term \(\sec^{-1}\left(\frac{5\sqrt{2}}{7}\right)\). We will convert this to an arctangent:
\[ \sec^{-1}(x) = \tan^{-1}\left(\sqrt{x^2 - 1}\right) \]
Calculating:
\[ \left(\frac{5\sqrt{2}}{7}\right)^2 - 1 = \frac{50}{49} - 1 = \frac{1}{49} \]
Thus,
\[ \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) = \tan^{-1}\left(\frac{1}{7}\right) \]

### Step 7: Combine all terms
Now we have:
\[ \tan^{-1}\left(\frac{507}{676}\right) + \tan^{-1}\left(\frac{1}{7}\right) \]
Using the sum formula again:
\[ \tan^{-1}\left(\frac{\frac{507}{676} + \frac{1}{7}}{1 - \frac{507}{676} \cdot \frac{1}{7}}\right) \]

Calculating the numerator:
\[ \frac{507 \cdot 7 + 676}{676 \cdot 7} = \frac{3549 + 676}{4732} = \frac{4225}{4732} \]

Calculating the denominator:
\[ 1 - \frac{507}{676 \cdot 7} = 1 - \frac{507}{4732} = \frac{4732 - 507}{4732} = \frac{4225}{4732} \]

Thus, we have:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer
The final result is:
\[ \frac{\pi}{4} \]