Solve the following equations :
`cos (tan^(-1) x ) = sin (cot^(-1). 3/4)`

Answer

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

### Step 1: Simplify the right side
Let \( \theta = \cot^{-1} \frac{3}{4} \). By the definition of cotangent, we have:
\[ \cot \theta = \frac{3}{4} \]
This means that the adjacent side is 3 and the opposite side is 4 in a right triangle. We can find the hypotenuse using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 2: Find \( \sin \theta \)
Now we can find \( \sin \theta \):
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} \]

### Step 3: Set up the equation
Now we have:
\[ \cos(\tan^{-1} x) = \frac{4}{5} \]

### Step 4: Simplify the left side
Let \( \phi = \tan^{-1} x \). By the definition of tangent, we have:
\[ \tan \phi = x \]
This means that in a right triangle, the opposite side is \( x \) and the adjacent side is 1. We can find the hypotenuse:
\[ \text{Hypotenuse} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 5: Find \( \cos \phi \)
Now we can find \( \cos \phi \):
\[ \cos \phi = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 6: Set up the equation
Now we have:
\[ \frac{1}{\sqrt{x^2 + 1}} = \frac{4}{5} \]

### Step 7: Cross-multiply and solve for \( x \)
Cross-multiplying gives:
\[ 5 = 4\sqrt{x^2 + 1} \]
Now, divide both sides by 4:
\[ \frac{5}{4} = \sqrt{x^2 + 1} \]
Now, square both sides:
\[ \left(\frac{5}{4}\right)^2 = x^2 + 1 \]
\[ \frac{25}{16} = x^2 + 1 \]
Now, subtract 1 from both sides:
\[ \frac{25}{16} - 1 = x^2 \]
Convert 1 to a fraction:
\[ \frac{25}{16} - \frac{16}{16} = x^2 \]
\[ \frac{9}{16} = x^2 \]

### Step 8: Solve for \( x \)
Taking the square root gives:
\[ x = \pm \frac{3}{4} \]

### Final Answer
Thus, the solutions for \( x \) are:
\[ x = \frac{3}{4} \quad \text{or} \quad x = -\frac{3}{4} \]

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