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 can follow these steps:

### Step 1: Assign Variables
Let:
\[ x = \sin^{-1}\left(\frac{5}{13}\right) \]
\[ y = \cos^{-1}\left(\frac{3}{5}\right) \]

### Step 2: Find \( \sin x \) and \( \cos y \)
From the definitions of inverse trigonometric functions, we have:
\[ \sin x = \frac{5}{13} \]
\[ \cos y = \frac{3}{5} \]

### Step 3: Find \( \cos x \) and \( \sin y \)
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
1. For \( \cos x \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \]
\[ \cos x = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

2. For \( \sin y \):
\[ \sin^2 y = 1 - \cos^2 y = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin y = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 4: Calculate \( \tan x \) and \( \tan y \)
Now, we can find:
\[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]
\[ \tan y = \frac{\sin y}{\cos y} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 5: Use the Formula for \( \tan(x + y) \)
Using the formula for the tangent of a sum:
\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
Substituting the values:
\[ \tan(x + y) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \left(\frac{5}{12} \cdot \frac{4}{3}\right)} \]

### Step 6: Simplify the Expression
1. Find a common denominator for the numerator:
\[ \frac{4}{3} = \frac{16}{12} \]
Thus,
\[ \tan(x + y) = \frac{\frac{5}{12} + \frac{16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \]

2. Simplify the denominator:
\[ 1 - \frac{5}{9} = \frac{4}{9} \]

3. Now, substituting back:
\[ \tan(x + y) = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21 \cdot 9}{12 \cdot 4} = \frac{189}{48} = \frac{63}{16} \]

### Step 7: Conclusion
Thus, we have:
\[ \tan(x + y) = \frac{63}{16} \]
Therefore, we can conclude:
\[ x + y = \tan^{-1}\left(\frac{63}{16}\right) \]
This implies:
\[ \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \]
Hence proved.