The valueof `2sin^(-1).(4)/(5)+2sin^(-1).(5)/(13)+2sin^(-1).(16)/(65)` is equal to

Options

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

Answer

To solve the expression \( 2\sin^{-1}\left(\frac{4}{5}\right) + 2\sin^{-1}\left(\frac{5}{13}\right) + 2\sin^{-1}\left(\frac{16}{65}\right) \), we can follow these steps:

### Step 1: Factor out the 2
We can factor out the 2 from the entire expression:
\[ 2\left(\sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right)\right) \]

### Step 2: Use the sine addition formula
We will use the formula for the sum of inverse sines:
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \]
Let \( x = \frac{4}{5} \) and \( y = \frac{5}{13} \).

### Step 3: Calculate \( \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) \)
First, we need to calculate:
- \( \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \)
- \( \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \)

Now substituting into the formula:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}\left(\frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5}\right) \]
Calculating the terms:
\[ \frac{4 \cdot 12}{65} + \frac{5 \cdot 3}{65} = \frac{48 + 15}{65} = \frac{63}{65} \]
Thus,
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}\left(\frac{63}{65}\right) \]

### Step 4: Combine with the third term
Now we need to add \( \sin^{-1}\left(\frac{63}{65}\right) + \sin^{-1}\left(\frac{16}{65}\right) \):
Using the same sine addition formula:
\[ \sin^{-1}\left(\frac{63}{65}\right) + \sin^{-1}\left(\frac{16}{65}\right) = \sin^{-1}\left(\frac{63}{65}\sqrt{1-\left(\frac{16}{65}\right)^2} + \frac{16}{65}\sqrt{1-\left(\frac{63}{65}\right)^2}\right) \]

Calculating:
- \( \sqrt{1 - \left(\frac{16}{65}\right)^2} = \sqrt{1 - \frac{256}{4225}} = \sqrt{\frac{3969}{4225}} = \frac{63}{65} \)
- \( \sqrt{1 - \left(\frac{63}{65}\right)^2} = \sqrt{1 - \frac{3969}{4225}} = \sqrt{\frac{256}{4225}} = \frac{16}{65} \)

Substituting:
\[ \sin^{-1}\left(\frac{63}{65} \cdot \frac{63}{65} + \frac{16}{65} \cdot \frac{16}{65}\right) = \sin^{-1}\left(\frac{3969 + 256}{4225}\right) = \sin^{-1}\left(\frac{4225}{4225}\right) = \sin^{-1}(1) = \frac{\pi}{2} \]

### Step 5: Final result
Thus, we have:
\[ 2\left(\sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right)\right) = 2 \cdot \frac{\pi}{2} = \pi \]

### Conclusion
The final value is:
\[ \pi \]