Prove that: `sin^(-1)(4/5)+sin^(-1)(5/(13))+sin^(-1)((16)/(65))=pi/2`

Explanation

To prove that \( \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right) = \frac{\pi}{2} \), we will follow these steps:

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