Prove that : `"tan"^(-1)3/4+"sin"^(-1)5/13="cos"^(-1)33/65`

Explanation

To prove that \( \tan^{-1}\left(\frac{3}{4}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right) \), we will follow these steps:

### Step 1: Convert \( \tan^{-1}\left(\frac{3}{4}\right) \) to cosine
Assume \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \). This means:
\[ \tan(\theta) = \frac{3}{4} \]
In a right triangle, the opposite side (perpendicular) is 3 and the adjacent side (base) is 4. We can find the hypotenuse \( h \) using the Pythagorean theorem:
\[ h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Now, we can find \( \cos(\theta) \):
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \]
Thus, we can write:
\[ \tan^{-1}\left(\frac{3}{4}\right) = \cos^{-1}\left(\frac{4}{5}\right) \]

### Step 2: Convert \( \sin^{-1}\left(\frac{5}{13}\right) \) to cosine
Let \( \phi = \sin^{-1}\left(\frac{5}{13}\right) \). This means:
\[ \sin(\phi) = \frac{5}{13} \]
In a right triangle, the opposite side (perpendicular) is 5 and the hypotenuse is 13. We can find the adjacent side (base) using the Pythagorean theorem:
\[ \text{adjacent} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \]
Now, we can find \( \cos(\phi) \):
\[ \cos(\phi) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13} \]
Thus, we can write:
\[ \sin^{-1}\left(\frac{5}{13}\right) = \cos^{-1}\left(\frac{12}{13}\right) \]

### Step 3: Combine the results
Now we have:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) \]

### Step 4: Use the cosine addition formula
Using the formula for the sum of cosines:
\[ \cos^{-1}(a) + \cos^{-1}(b) = \cos^{-1}(ab - \sqrt{(1-a^2)(1-b^2)}) \]
Let \( a = \frac{4}{5} \) and \( b = \frac{12}{13} \):
1. Calculate \( ab \):
\[ ab = \frac{4}{5} \cdot \frac{12}{13} = \frac{48}{65} \]
2. Calculate \( 1 - a^2 \):
\[ 1 - a^2 = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
3. Calculate \( 1 - b^2 \):
\[ 1 - b^2 = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \]
4. Calculate the square root:
\[ \sqrt{(1-a^2)(1-b^2)} = \sqrt{\frac{9}{25} \cdot \frac{25}{169}} = \sqrt{\frac{9}{169}} = \frac{3}{13} \]

### Step 5: Combine to find the final cosine value
Now substituting back:
\[ \cos^{-1}\left(\frac{48}{65} - \frac{3}{13}\right) \]
Convert \( \frac{3}{13} \) to have a common denominator:
\[ \frac{3}{13} = \frac{15}{65} \]
Thus:
\[ \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \]
So we have:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right) \]

### Conclusion
Therefore, we have proved that:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right) \]