The value of `sec ( 2 cot^(-1) 2 + cos^(-1) . 3/5) ` is equal to

Options

A. `25/24`
B. ` - 24/7`
C. `25/7`
D. `- 25/7`

Answer

To find the value of \( \sec(2 \cot^{-1}(2) + \cos^{-1}(\frac{3}{5})) \), we can follow these steps:

### Step 1: Convert \( \cot^{-1}(2) \) to \( \tan^{-1} \)
Let \( x = \cot^{-1}(2) \). Then, we have:
\[ \cot(x) = 2 \implies \tan(x) = \frac{1}{\cot(x)} = \frac{1}{2} \]
Thus, we can write:
\[ x = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 2: Calculate \( 2 \cot^{-1}(2) \)
Now we need to find \( 2x \):
\[ 2x = 2 \tan^{-1}\left(\frac{1}{2}\right) \]
Using the double angle formula for tangent:
\[ \tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)} = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]
Thus, we have:
\[ 2 \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 3: Convert \( \cos^{-1}(\frac{3}{5}) \) to \( \sin^{-1} \)
Let \( y = \cos^{-1}\left(\frac{3}{5}\right) \). Then:
\[ \cos(y) = \frac{3}{5} \]
To find \( \sin(y) \), we use the Pythagorean identity:
\[ \sin^2(y) + \cos^2(y) = 1 \implies \sin^2(y) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \implies \sin(y) = \frac{4}{5} \]

### Step 4: Calculate \( \tan(y) \)
Now, we can find \( \tan(y) \):
\[ \tan(y) = \frac{\sin(y)}{\cos(y)} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 5: Combine the angles
Now, we need to find:
\[ \sec(2 \cot^{-1}(2) + \cos^{-1}(\frac{3}{5})) = \sec\left(\tan^{-1}\left(\frac{4}{3}\right) + \cos^{-1}\left(\frac{3}{5}\right)\right) \]
Using the formula for the tangent of the sum of angles:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Here, \( A = \tan^{-1}\left(\frac{4}{3}\right) \) and \( B = \cos^{-1}\left(\frac{3}{5}\right) \):
\[ \tan(A) = \frac{4}{3}, \quad \tan(B) = \frac{4}{3} \]
Thus:
\[ \tan(A + B) = \frac{\frac{4}{3} + \frac{4}{3}}{1 - \left(\frac{4}{3}\right)\left(\frac{4}{3}\right)} = \frac{\frac{8}{3}}{1 - \frac{16}{9}} = \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{8 \cdot 9}{3 \cdot 7} = -\frac{24}{7} \]

### Step 6: Find \( \sec(A + B) \)
Now, we find:
\[ \sec(A + B) = \frac{1}{\cos(A + B)} = \sqrt{1 + \tan^2(A + B)} = \sqrt{1 + \left(-\frac{24}{7}\right)^2} = \sqrt{1 + \frac{576}{49}} = \sqrt{\frac{625}{49}} = \frac{25}{7} \]

### Final Answer
Thus, the value of \( \sec(2 \cot^{-1}(2) + \cos^{-1}(\frac{3}{5})) \) is:
\[ \frac{25}{7} \]