What is the value of `sin^(-1)""(4)/(5)+2tan^(-1)""(1)/(3)`?

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Answer

To find the value of \( \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) \), we can follow these steps:

### Step 1: Write the expression
We start with the expression:
\[ \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Use the formula for \( 2\tan^{-1}(x) \)
We know that:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
In our case, let \( x = \frac{1}{3} \). Plugging this into the formula gives:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) \]

### Step 3: Calculate the numerator and denominator
Calculating the numerator:
\[ 2 \cdot \frac{1}{3} = \frac{2}{3} \]
Calculating the denominator:
\[ 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus, we have:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) = \tan^{-1}\left(\frac{2 \cdot 9}{3 \cdot 8}\right) = \tan^{-1}\left(\frac{6}{8}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 4: Substitute back into the expression
Now substituting back, we have:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 5: Use the identity for \( \sin^{-1}(x) + \tan^{-1}(y) \)
We can use the identity:
\[ \tan^{-1}(y) = \cos^{-1}\left(\frac{1}{\sqrt{1+y^2}}\right) \]
Thus,
\[ \tan^{-1}\left(\frac{3}{4}\right) = \cos^{-1}\left(\frac{1}{\sqrt{1+\left(\frac{3}{4}\right)^2}}\right) \]

### Step 6: Calculate \( \cos^{-1} \)
Calculating \( 1 + \left(\frac{3}{4}\right)^2 \):
\[ 1 + \frac{9}{16} = \frac{16}{16} + \frac{9}{16} = \frac{25}{16} \]
So,
\[ \sqrt{1 + \left(\frac{3}{4}\right)^2} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]
Thus,
\[ \tan^{-1}\left(\frac{3}{4}\right) = \cos^{-1}\left(\frac{4}{5}\right) \]

### Step 7: Combine the results
Now we have:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) \]
Using the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
So,
\[ \sin^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2} \]

### Final Answer
Thus, the value of \( \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) \) is:
\[ \frac{\pi}{2} \]