Prove that : ` sin^(-1)(5/13) +cos ^(−1)(3/5)=tan (−1)(63/16)` ​

Answer

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: Set Up the Equations
Let:
\[ x = \sin^{-1}\left(\frac{5}{13}\right) \]
Then, we have:
\[ \sin x = \frac{5}{13} \]

### Step 2: Find the Cosine of \( x \)
Using the Pythagorean identity:
\[ \cos^2 x + \sin^2 x = 1 \]
Substituting \( \sin x \):
\[ \cos^2 x + \left(\frac{5}{13}\right)^2 = 1 \]
\[ \cos^2 x + \frac{25}{169} = 1 \]
\[ \cos^2 x = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \]
Thus,
\[ \cos x = \frac{12}{13} \]

### Step 3: Set Up the Second Equation
Now, let:
\[ y = \cos^{-1}\left(\frac{3}{5}\right) \]
Then, we have:
\[ \cos y = \frac{3}{5} \]

### Step 4: Find the Sine of \( y \)
Using the Pythagorean identity again:
\[ \sin^2 y + \cos^2 y = 1 \]
Substituting \( \cos y \):
\[ \sin^2 y + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 y + \frac{9}{25} = 1 \]
\[ \sin^2 y = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} \]
Thus,
\[ \sin y = \frac{4}{5} \]

### Step 5: Combine the Angles
Now we need to find:
\[ x + y = \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \]

### Step 6: Use the Tangent Addition Formula
We can express \( \tan(x + y) \) using the tangent addition formula:
\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]

Calculating \( \tan x \) and \( \tan y \):
\[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]
\[ \tan y = \frac{\sin y}{\cos y} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 7: Substitute into the Tangent Addition Formula
Now substituting into the formula:
\[ \tan(x + y) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \left(\frac{5}{12} \cdot \frac{4}{3}\right)} \]

Finding a common denominator for the numerator:
\[ \tan(x + y) = \frac{\frac{5}{12} + \frac{16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \]

Calculating the denominator:
\[ 1 - \frac{5}{9} = \frac{9 - 5}{9} = \frac{4}{9} \]

Thus,
\[ \tan(x + y) = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21 \cdot 9}{12 \cdot 4} = \frac{189}{48} = \frac{63}{16} \]

### Conclusion
Since \( \tan(x + y) = \frac{63}{16} \), we have:
\[ x + y = \tan^{-1}\left(\frac{63}{16}\right) \]
Thus, we have proved that:
\[ \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \]