Let `alpha = 3 cos^(-1) (5/sqrt28) + 3 tan^(-1) ( sqrt3/2) " and " beta = 4 sin ^(-1) ((7sqrt2)/10) - 4 tan^(-1) (3/4)`, then which of the following does not hold (s) good ?

Options

A. `alpha lt pi " but " beta gt pi`
B. `alpha gt pi " but " beta lt pi`
C. Both `alpha " and " beta` are equal
D. `cos ( alpha + beta) = 0`

Answer

To solve the problem, we need to evaluate the expressions for \(\alpha\) and \(\beta\) and determine which of the given options does not hold true.

### Step 1: Calculate \(\alpha\)

Given:
\[ \alpha = 3 \cos^{-1} \left(\frac{5}{\sqrt{28}}\right) + 3 \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \]

Let:
\[ \cos^{-1} \left(\frac{5}{\sqrt{28}}\right) = a \]
Then:
\[ \cos a = \frac{5}{\sqrt{28}} \]

### Step 2: Construct a right triangle for \(a\)

Using the cosine definition:
- Hypotenuse = \(\sqrt{28}\)
- Base = 5
- Perpendicular = \(\sqrt{28^2 - 5^2} = \sqrt{28 - 25} = \sqrt{3}\)

### Step 3: Calculate \(\tan a\)

From the triangle:
\[ \tan a = \frac{\text{Perpendicular}}{\text{Base}} = \frac{\sqrt{3}}{5} \]
Thus:
\[ a = \tan^{-1} \left(\frac{\sqrt{3}}{5}\right) \]

### Step 4: Substitute back into \(\alpha\)

Now substituting \(a\) back into the expression for \(\alpha\):
\[ \alpha = 3 \tan^{-1} \left(\frac{\sqrt{3}}{5}\right) + 3 \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \]

### Step 5: Use the formula for \(\tan^{-1}(A) + \tan^{-1}(B)\)

Using the formula:
\[ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A + B}{1 - AB}\right) \]
Let:
\[ A = \frac{\sqrt{3}}{5}, \quad B = \frac{\sqrt{3}}{2} \]

Calculating:
\[ AB = \frac{\sqrt{3}}{5} \cdot \frac{\sqrt{3}}{2} = \frac{3}{10} \]
Then:
\[ A + B = \frac{\sqrt{3}}{5} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3} + 5\sqrt{3}}{10} = \frac{7\sqrt{3}}{10} \]

Thus:
\[ \alpha = 3 \tan^{-1} \left(\frac{\frac{7\sqrt{3}}{10}}{1 - \frac{3}{10}}\right) = 3 \tan^{-1} \left(\frac{7\sqrt{3}}{7}\right) = 3 \tan^{-1} \left(\sqrt{3}\right) \]

Since \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\):
\[ \alpha = 3 \cdot \frac{\pi}{3} = \pi \]

### Step 6: Calculate \(\beta\)

Given:
\[ \beta = 4 \sin^{-1} \left(\frac{7\sqrt{2}}{10}\right) - 4 \tan^{-1} \left(\frac{3}{4}\right) \]

Let:
\[ \sin^{-1} \left(\frac{7\sqrt{2}}{10}\right) = b \]
Then:
\[ \sin b = \frac{7\sqrt{2}}{10} \]

### Step 7: Construct a right triangle for \(b\)

Using the sine definition:
- Hypotenuse = 10
- Perpendicular = \(7\sqrt{2}\)
- Base = \(\sqrt{10^2 - (7\sqrt{2})^2} = \sqrt{100 - 98} = \sqrt{2}\)

### Step 8: Calculate \(\tan b\)

From the triangle:
\[ \tan b = \frac{\text{Perpendicular}}{\text{Base}} = \frac{7\sqrt{2}}{\sqrt{2}} = 7 \]
Thus:
\[ b = \tan^{-1}(7) \]

### Step 9: Substitute back into \(\beta\)

Now substituting \(b\) back into the expression for \(\beta\):
\[ \beta = 4 \tan^{-1}(7) - 4 \tan^{-1} \left(\frac{3}{4}\right) \]

### Step 10: Use the formula for \(\tan^{-1}(A) - \tan^{-1}(B)\)

Using the formula:
\[ \tan^{-1} A - \tan^{-1} B = \tan^{-1} \left(\frac{A - B}{1 + AB}\right) \]

Calculating:
\[ A = 7, \quad B = \frac{3}{4} \]
Then:
\[ AB = 7 \cdot \frac{3}{4} = \frac{21}{4} \]
Thus:
\[ \beta = 4 \tan^{-1} \left(\frac{7 - \frac{3}{4}}{1 + \frac{21}{4}}\right) = 4 \tan^{-1} \left(\frac{\frac{28}{4} - \frac{3}{4}}{\frac{25}{4}}\right) = 4 \tan^{-1 \left(\frac{25}{25}\right)} = 4 \tan^{-1}(1) \]

Since \(\tan^{-1}(1) = \frac{\pi}{4}\):
\[ \beta = 4 \cdot \frac{\pi}{4} = \pi \]

### Step 11: Compare \(\alpha\) and \(\beta\)

We found:
\[ \alpha = \pi \quad \text{and} \quad \beta = \pi \]

### Conclusion

Now we check the options:
1. \(\alpha < \pi\) but \(\beta > \pi\) - Incorrect
2. \(\alpha > \pi\) but \(\beta < \pi\) - Incorrect
3. \(\alpha = \beta\) - Correct
4. \(\cos(\alpha + \beta) = 0\) - Incorrect since \(\alpha + \beta = 2\pi\) and \(\cos(2\pi) = 1\)

Thus, the option that does not hold true is option 4.