Consider the following statements
1. `sin^(-1)""(4)/(5)+sin^(-1)""(3)/(5)=(pi)/(2)`
2. `tan^(-1)sqrt3+tan^(-1)1=-tan^(-1)(2+sqrt3)`
Which of the above statements (s) is/are correct ?

Options

A. Only 1
B. Only 2
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To determine the correctness of the given statements, we will analyze each statement step by step.

### Statement 1:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \frac{\pi}{2} \]

**Step 1: Use the formula for the sum of inverse sine functions.**
The formula for the sum of two inverse sine functions is:
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \]
where \( x = \frac{4}{5} \) and \( y = \frac{3}{5} \).

**Step 2: Calculate \( \sqrt{1 - y^2} \) and \( \sqrt{1 - x^2} \).**
\[ \sqrt{1 - y^2} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
\[ \sqrt{1 - x^2} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

**Step 3: Substitute back into the formula.**
Now substituting into the formula:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{4}{5} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{3}{5}\right) = \sin^{-1}\left(\frac{16}{25} + \frac{9}{25}\right) = \sin^{-1}\left(\frac{25}{25}\right) = \sin^{-1}(1) \]

**Step 4: Conclusion for Statement 1.**
Since \(\sin^{-1}(1) = \frac{\pi}{2}\), the first statement is correct.

### Statement 2:
\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = -\tan^{-1}(2 + \sqrt{3}) \]

**Step 1: Use the formula for the sum of inverse tangent functions.**
The formula for the sum of two inverse tangent functions is:
\[ \tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \]
where \( A = \sqrt{3} \) and \( B = 1 \).

**Step 2: Substitute \( A \) and \( B \) into the formula.**
\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = \tan^{-1}\left(\frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1}\right) = \tan^{-1}\left(\frac{\sqrt{3} + 1}{1 - \sqrt{3}}\right) \]

**Step 3: Simplify the expression.**
Rationalizing the denominator:
\[ \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{3 + \sqrt{3} + \sqrt{3} + 1}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3} \]

**Step 4: Conclusion for Statement 2.**
Thus, we have:
\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = \tan^{-1}(-2 - \sqrt{3}) = -\tan^{-1}(2 + \sqrt{3}) \]
This shows that the second statement is also correct.

### Final Conclusion:
Both statements are correct.