`tan[cos^(-1) ""4/5 + tan^(-1)"" 2/3] " or " tan[sin^(-1) ""3/5+ cot^(-1)""3/2]=`

Options

A. `6/17`
B. `7/16`
C. `17/6`
D. none

Answer

To solve the problem \( \tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) \) and \( \tan\left(\sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right)\right) \), we can follow these steps:

### Step 1: Solve \( \tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) \)

Let \( \alpha = \cos^{-1}\left(\frac{4}{5}\right) \) and \( \beta = \tan^{-1}\left(\frac{2}{3}\right) \).

Using the identity for tangent of a sum:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

### Step 2: Find \( \tan \alpha \)

From \( \alpha = \cos^{-1}\left(\frac{4}{5}\right) \):
- We know that \( \cos \alpha = \frac{4}{5} \).
- To find \( \sin \alpha \), we use the Pythagorean identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \implies \sin^2 \alpha = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
\[ \sin \alpha = \frac{3}{5} \]
Thus,
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 3: Find \( \tan \beta \)

From \( \beta = \tan^{-1}\left(\frac{2}{3}\right) \):
\[ \tan \beta = \frac{2}{3} \]

### Step 4: Substitute into the tangent sum formula

Now substituting into the formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right)} \]

### Step 5: Simplify the numerator and denominator

Finding a common denominator for the numerator:
\[ \tan \alpha + \tan \beta = \frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12} \]

Now for the denominator:
\[ 1 - \tan \alpha \tan \beta = 1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Final calculation

Now substituting back:
\[ \tan(\alpha + \beta) = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \cdot 2 = \frac{17}{6} \]

### Step 7: Solve \( \tan\left(\sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right)\right) \)

Let \( \gamma = \sin^{-1}\left(\frac{3}{5}\right) \) and \( \delta = \cot^{-1}\left(\frac{3}{2}\right) \).

Using the same tangent sum formula:
\[ \tan(\gamma + \delta) = \frac{\tan \gamma + \tan \delta}{1 - \tan \gamma \tan \delta} \]

### Step 8: Find \( \tan \gamma \)

From \( \gamma = \sin^{-1}\left(\frac{3}{5}\right) \):
- We know that \( \sin \gamma = \frac{3}{5} \).
- To find \( \cos \gamma \):
\[ \cos^2 \gamma + \sin^2 \gamma = 1 \implies \cos^2 \gamma = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \cos \gamma = \frac{4}{5} \]
Thus,
\[ \tan \gamma = \frac{\sin \gamma}{\cos \gamma} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 9: Find \( \tan \delta \)

From \( \delta = \cot^{-1}\left(\frac{3}{2}\right) \):
\[ \tan \delta = \frac{1}{\cot \delta} = \frac{2}{3} \]

### Step 10: Substitute into the tangent sum formula

Now substituting into the formula:
\[ \tan(\gamma + \delta) = \frac{\tan \gamma + \tan \delta}{1 - \tan \gamma \tan \delta} = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right)} \]

### Step 11: Simplify the numerator and denominator

Finding a common denominator for the numerator:
\[ \tan \gamma + \tan \delta = \frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12} \]

Now for the denominator:
\[ 1 - \tan \gamma \tan \delta = 1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 12: Final calculation

Now substituting back:
\[ \tan(\gamma + \delta) = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \cdot 2 = \frac{17}{6} \]

### Conclusion

Both expressions yield the same result:
\[ \tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) = \tan\left(\sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right)\right) = \frac{17}{6} \]