The numerical value of `cot(2sin^(-1)\ 3/5+cos^(-1)\ 3/5)` is

Options

A. `(-4)/3`
B. `(-3)/4`
C. `3/4`
D. `4/3`

Explanation

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

### Step 1: Define the angle
Let \( \theta = \sin^{-1}(3/5) \). Therefore, we have:
\[ 2\sin^{-1}(3/5) = 2\theta \]

### Step 2: Find \( \cos^{-1}(3/5) \)
Using the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), we can express \( \cos^{-1}(3/5) \) as:
\[ \cos^{-1}(3/5) = \frac{\pi}{2} - \theta \]

### Step 3: Substitute into the cotangent expression
Now, we substitute \( 2\theta \) and \( \frac{\pi}{2} - \theta \) into the cotangent:
\[ \cot(2\theta + \cos^{-1}(3/5)) = \cot(2\theta + \frac{\pi}{2} - \theta) = \cot(\theta + \frac{\pi}{2}) \]

### Step 4: Use cotangent identity
Using the cotangent identity \( \cot(x + \frac{\pi}{2}) = -\tan(x) \), we have:
\[ \cot(\theta + \frac{\pi}{2}) = -\tan(\theta) \]

### Step 5: Find \( \tan(\theta) \)
Since \( \theta = \sin^{-1}(3/5) \), we can find \( \tan(\theta) \) using the sine and cosine values. We know:
\[ \sin(\theta) = \frac{3}{5} \]
To find \( \cos(\theta) \), we use the Pythagorean theorem:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus, we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{4/5} = \frac{3}{4} \]

### Step 6: Substitute back to find cotangent
Now substituting back, we get:
\[ -\tan(\theta) = -\frac{3}{4} \]

### Final Answer
Thus, the numerical value of \( \cot(2\sin^{-1}(3/5) + \cos^{-1}(3/5)) \) is:
\[ \boxed{-\frac{3}{4}} \]