Show that `sin^(-1)(5/13)+cos^(-1)(3/5)=tan^(-1)(63/16)`.

Explanation

To prove that \( \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \), we will follow these steps:

### Step 1: Define the angles
Let:
\[ \theta = \sin^{-1}\left(\frac{5}{13}\right) \]
\[ \phi = \cos^{-1}\left(\frac{3}{5}\right) \]

### Step 2: Find the sides of the triangle for \( \theta \)
From \( \theta = \sin^{-1}\left(\frac{5}{13}\right) \):
- Opposite side (perpendicular) = 5
- Hypotenuse = 13

Using the Pythagorean theorem, we can find the adjacent side (base):
\[ \text{Adjacent} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \]

### Step 3: Find the sides of the triangle for \( \phi \)
From \( \phi = \cos^{-1}\left(\frac{3}{5}\right) \):
- Adjacent side (base) = 3
- Hypotenuse = 5

Using the Pythagorean theorem, we can find the opposite side (perpendicular):
\[ \text{Opposite} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]

### Step 4: Calculate \( \tan(\theta) \) and \( \tan(\phi) \)
Now we can find \( \tan(\theta) \) and \( \tan(\phi) \):
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12} \]
\[ \tan(\phi) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{3} \]

### Step 5: Use the tangent addition formula
We will use the formula for the tangent of the sum of two angles:
\[ \tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta) \tan(\phi)} \]
Substituting the values:
\[ \tan(\theta + \phi) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \left(\frac{5}{12} \cdot \frac{4}{3}\right)} \]

### Step 6: Simplify the numerator
To add the fractions in the numerator:
\[ \frac{4}{3} = \frac{16}{12} \]
Thus,
\[ \tan(\theta + \phi) = \frac{\frac{5}{12} + \frac{16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \]

### Step 7: Simplify the denominator
Calculating the denominator:
\[ 1 - \frac{5}{9} = \frac{4}{9} \]
So,
\[ \tan(\theta + \phi) = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21 \cdot 9}{12 \cdot 4} = \frac{189}{48} \]

### Step 8: Simplify the fraction
Now simplifying \( \frac{189}{48} \):
\[ \frac{189}{48} = \frac{63}{16} \]

### Step 9: Final result
Thus, we have:
\[ \tan(\theta + \phi) = \frac{63}{16} \]
This implies:
\[ \theta + \phi = \tan^{-1}\left(\frac{63}{16}\right) \]
Therefore, we conclude that:
\[ \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \]