`cos(tan^(-1)3/4)+cos(tan^(-1)x)` is equal to

Options

A. `4/5+x/sqrt(1+x^2)`
B. `3/5+1/sqrt(1+x^2)`
C. `4/5+1/sqrt(1+x^2)`
D. `3/5+x/sqrt(1+x^2)`

Answer

To solve the expression \( \cos(\tan^{-1}(3/4)) + \cos(\tan^{-1}(x)) \), we will follow these steps:

### Step 1: Define the angles
Let \( \theta = \tan^{-1}(3/4) \). This means that:
\[ \tan(\theta) = \frac{3}{4} \]

### Step 2: Create a right triangle
From the definition of tangent, we can visualize a right triangle where the opposite side is 3 and the adjacent side is 4.

### Step 3: Calculate the hypotenuse
Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Find \( \cos(\theta) \)
Now, we can find \( \cos(\theta) \):
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \]

### Step 5: Define the second angle
Let \( \alpha = \tan^{-1}(x) \). This means that:
\[ \tan(\alpha) = x \]

### Step 6: Create another right triangle
For this angle, we can visualize another right triangle where the opposite side is \( x \) and the adjacent side is 1.

### Step 7: Calculate the hypotenuse for the second triangle
Using the Pythagorean theorem again:
\[ \text{Hypotenuse} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 8: Find \( \cos(\alpha) \)
Now we can find \( \cos(\alpha) \):
\[ \cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 9: Combine the results
Now we can combine the results:
\[ \cos(\tan^{-1}(3/4)) + \cos(\tan^{-1}(x)) = \frac{4}{5} + \frac{1}{\sqrt{x^2 + 1}} \]

### Final Result
Thus, the expression \( \cos(\tan^{-1}(3/4)) + \cos(\tan^{-1}(x)) \) simplifies to:
\[ \frac{4}{5} + \frac{1}{\sqrt{x^2 + 1}} \]