`sec^(2) (tan^(-1) 4) + cosec^(2) (cot ^(-1) 3) =?`

Options

A. 30
B. 29
C. 27
D. 25

Answer

To solve the problem \( \sec^2(\tan^{-1} 4) + \csc^2(\cot^{-1} 3) \), we will follow these steps:

### Step 1: Rewrite the expressions using inverse trigonometric functions
We start with the expression:
\[ \sec^2(\tan^{-1} 4) + \csc^2(\cot^{-1} 3) \]

### Step 2: Let \( \theta = \tan^{-1}(4) \)
From the definition of the tangent function, we have:
\[ \tan(\theta) = 4 \implies \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{1} \]
Using the Pythagorean theorem, we can find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \]

### Step 3: Calculate \( \sec^2(\theta) \)
Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \) and \( \cos(\theta) = \frac{1}{\sqrt{1 + \tan^2(\theta)}} \):
\[ \sec^2(\theta) = 1 + \tan^2(\theta) = 1 + 4^2 = 1 + 16 = 17 \]

### Step 4: Let \( \phi = \cot^{-1}(3) \)
From the definition of the cotangent function, we have:
\[ \cot(\phi) = 3 \implies \frac{\text{adjacent}}{\text{opposite}} = \frac{3}{1} \]
Using the Pythagorean theorem, we can find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 5: Calculate \( \csc^2(\phi) \)
Since \( \csc(\phi) = \frac{1}{\sin(\phi)} \) and \( \sin(\phi) = \frac{1}{\sqrt{1 + \cot^2(\phi)}} \):
\[ \csc^2(\phi) = 1 + \cot^2(\phi) = 1 + 3^2 = 1 + 9 = 10 \]

### Step 6: Combine the results
Now we can combine the results from Steps 3 and 5:
\[ \sec^2(\tan^{-1} 4) + \csc^2(\cot^{-1} 3) = 17 + 10 = 27 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{27} \]