Show that: `tan(1/2sin^(-1)(3/4))=(4\- sqrt(7))/3`

Explanation

To show that \( \tan\left(\frac{1}{2} \sin^{-1}\left(\frac{3}{4}\right)\right) = \frac{4 - \sqrt{7}}{3} \), we will follow these steps:

### Step 1: Set up the problem
Let \( \theta = \sin^{-1}\left(\frac{3}{4}\right) \). This implies that \( \sin(\theta) = \frac{3}{4} \).

### Step 2: Find \( \cos(\theta) \)
Using the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
We can find \( \cos(\theta) \):
\[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16} \]
Thus,
\[ \cos(\theta) = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 3: Use the half-angle formula for tangent
We will use the half-angle formula for tangent:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \]
Substituting the values we found:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} = \frac{\frac{3}{4}}{1 + \frac{\sqrt{7}}{4}} \]

### Step 4: Simplify the expression
Now, simplify the denominator:
\[ 1 + \frac{\sqrt{7}}{4} = \frac{4 + \sqrt{7}}{4} \]
Thus, we have:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\frac{3}{4}}{\frac{4 + \sqrt{7}}{4}} = \frac{3}{4 + \sqrt{7}} \]

### Step 5: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{3(4 - \sqrt{7})}{(4 + \sqrt{7})(4 - \sqrt{7})} \]
Calculating the denominator:
\[ (4 + \sqrt{7})(4 - \sqrt{7}) = 4^2 - (\sqrt{7})^2 = 16 - 7 = 9 \]
Thus, we have:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{3(4 - \sqrt{7})}{9} = \frac{4 - \sqrt{7}}{3} \]

### Conclusion
We have shown that:
\[ \tan\left(\frac{1}{2} \sin^{-1}\left(\frac{3}{4}\right)\right) = \frac{4 - \sqrt{7}}{3} \]