`sin(sin^(-1)((1)/(3))+sec^(-1)(3))+cos(tan^(-1)(1/2)+tan^(-1)2)`=

Options

A. `1`
B. `2`
C. `0`
D. `-1`

Answer

To solve the expression \( \sin(\sin^{-1}(\frac{1}{3}) + \sec^{-1}(3)) + \cos(\tan^{-1}(\frac{1}{2}) + \tan^{-1}(2)) \), we can break it down step by step.

### Step 1: Simplify \( \sec^{-1}(3) \)
Recall that \( \sec^{-1}(x) = \cos^{-1}(\frac{1}{x}) \). Therefore,
\[ \sec^{-1}(3) = \cos^{-1}(\frac{1}{3}). \]

### Step 2: Rewrite the first part
Now, we can rewrite the first part of the expression:
\[ \sin(\sin^{-1}(\frac{1}{3}) + \sec^{-1}(3)) = \sin(\sin^{-1}(\frac{1}{3}) + \cos^{-1}(\frac{1}{3})). \]

### Step 3: Use the identity for sine
Using the identity \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), where \( a = \sin^{-1}(\frac{1}{3}) \) and \( b = \cos^{-1}(\frac{1}{3}) \):
- \( \sin(\sin^{-1}(\frac{1}{3})) = \frac{1}{3} \)
- \( \cos(\sin^{-1}(\frac{1}{3})) = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \)
- \( \sin(\cos^{-1}(\frac{1}{3})) = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \frac{2\sqrt{2}}{3} \)
- \( \cos(\cos^{-1}(\frac{1}{3})) = \frac{1}{3} \)

Now substituting these values:
\[ \sin(\sin^{-1}(\frac{1}{3}) + \cos^{-1}(\frac{1}{3})) = \frac{1}{3} \cdot \frac{1}{3} + \frac{2\sqrt{2}}{3} \cdot \frac{2\sqrt{2}}{3} = \frac{1}{9} + \frac{8}{9} = 1. \]

### Step 4: Simplify \( \tan^{-1}(\frac{1}{2}) + \tan^{-1}(2) \)
Using the identity \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) when \( xy < 1 \):
\[ \tan^{-1}(\frac{1}{2}) + \tan^{-1}(2) = \tan^{-1}\left(\frac{\frac{1}{2} + 2}{1 - \frac{1}{2} \cdot 2}\right) = \tan^{-1}\left(\frac{\frac{5}{2}}{0}\right) = \frac{\pi}{2}. \]

### Step 5: Evaluate \( \cos(\tan^{-1}(\frac{1}{2}) + \tan^{-1}(2)) \)
Since \( \tan^{-1}(\frac{1}{2}) + \tan^{-1}(2) = \frac{\pi}{2} \):
\[ \cos\left(\frac{\pi}{2}\right) = 0. \]

### Step 6: Combine results
Now we can combine the results:
\[ \sin(\sin^{-1}(\frac{1}{3}) + \sec^{-1}(3)) + \cos(\tan^{-1}(\frac{1}{2}) + \tan^{-1}(2)) = 1 + 0 = 1. \]

### Final Answer
Thus, the final answer is:
\[ \boxed{1}. \]