If: `cos(tan^(-1)x)=sin (cot^(-1).(3)/(4)),` then : x =

Options

A. `1/2`
B. `3/4`
C. `pm(5)/(6)`
D. none of these

Answer

To solve the equation \( \cos(\tan^{-1} x) = \sin(\cot^{-1} \frac{3}{4}) \), we can follow these steps:

### Step 1: Rewrite the right-hand side
We know that:
\[ \cot^{-1} \theta = \tan^{-1} \left(\frac{1}{\theta}\right) \]
Thus, we can rewrite \( \cot^{-1} \frac{3}{4} \) as:
\[ \cot^{-1} \frac{3}{4} = \tan^{-1} \left(\frac{4}{3}\right) \]
So, we have:
\[ \sin(\cot^{-1} \frac{3}{4}) = \sin\left(\tan^{-1} \frac{4}{3}\right) \]

### Step 2: Set up triangles
Let \( \theta_1 = \tan^{-1} x \) and \( \theta_2 = \tan^{-1} \frac{4}{3} \).

For \( \theta_1 \):
- Opposite side = \( x \)
- Adjacent side = \( 1 \)
- Hypotenuse = \( \sqrt{1 + x^2} \)

For \( \theta_2 \):
- Opposite side = \( 4 \)
- Adjacent side = \( 3 \)
- Hypotenuse = \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)

### Step 3: Find cosine and sine values
Now, we can find:
\[ \cos(\theta_1) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{1 + x^2}} \]
\[ \sin(\theta_2) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} \]

### Step 4: Set up the equation
Now we can set up the equation:
\[ \frac{1}{\sqrt{1 + x^2}} = \frac{4}{5} \]

### Step 5: Cross-multiply and solve for \( x^2 \)
Cross-multiplying gives:
\[ 5 = 4\sqrt{1 + x^2} \]
Squaring both sides:
\[ 25 = 16(1 + x^2) \]
Expanding this:
\[ 25 = 16 + 16x^2 \]
Subtracting 16 from both sides:
\[ 9 = 16x^2 \]
Dividing both sides by 16:
\[ x^2 = \frac{9}{16} \]

### Step 6: Take the square root
Taking the square root of both sides gives:
\[ x = \pm \frac{3}{4} \]
Since we are looking for the positive value, we have:
\[ x = \frac{3}{4} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{3}{4}} \]