`sin^(-1) ""12/13 + cos^(-1)"" 4/5 tan^(-1) ""63/16=`

Options

A. 0
B. `pi//2`
C. `pi`
D. `3pi//2`

Answer

To solve the problem \( \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) \), we will break it down step by step.

### Step 1: Define the angles
Let:
- \( \alpha = \sin^{-1}\left(\frac{12}{13}\right) \)
- \( \beta = \cos^{-1}\left(\frac{4}{5}\right) \)
- \( \gamma = \tan^{-1}\left(\frac{63}{16}\right) \)

### Step 2: Find \( \sin \alpha \) and \( \cos \beta \)
From the definitions:
- \( \sin \alpha = \frac{12}{13} \)
- \( \cos \beta = \frac{4}{5} \)

### Step 3: Calculate \( \cos \alpha \) and \( \sin \beta \)
Using the Pythagorean identity:
- For \( \alpha \):
\[ \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

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

### Step 4: Find \( \tan \alpha \) and \( \tan \beta \)
Now, we can find:
- \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \)
- \( \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

### Step 5: Use the tangent addition formula
We need to find \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - \left(\frac{12}{5} \cdot \frac{3}{4}\right)} \]

Calculating the numerator:
\[ \frac{12}{5} + \frac{3}{4} = \frac{48}{20} + \frac{15}{20} = \frac{63}{20} \]

Calculating the denominator:
\[ 1 - \left(\frac{12 \cdot 3}{5 \cdot 4}\right) = 1 - \frac{36}{20} = \frac{20 - 36}{20} = \frac{-16}{20} = -\frac{4}{5} \]

Thus:
\[ \tan(\alpha + \beta) = \frac{\frac{63}{20}}{-\frac{4}{5}} = \frac{63}{20} \cdot -\frac{5}{4} = -\frac{315}{80} \]

### Step 6: Add \( \tan^{-1} \left(\frac{63}{16}\right) \)
Now we need to calculate:
\[ \tan(\alpha + \beta + \gamma) = \tan\left(\tan^{-1}\left(-\frac{315}{80}\right) + \tan^{-1}\left(\frac{63}{16}\right)\right) \]

Using the tangent addition formula again:
\[ \tan(\alpha + \beta + \gamma) = \frac{-\frac{315}{80} + \frac{63}{16}}{1 - \left(-\frac{315}{80} \cdot \frac{63}{16}\right)} \]

Calculating the numerator:
\[ -\frac{315}{80} + \frac{63}{16} = -\frac{315}{80} + \frac{315}{80} = 0 \]

### Final Result
Thus, we have:
\[ \tan(\alpha + \beta + \gamma) = 0 \]
This implies:
\[ \alpha + \beta + \gamma = \tan^{-1}(0) = 0 \]

So the final answer is:
\[ \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = 0 \]