Consider the following :
1. ${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{3}{5}=\frac{\pi }{2}$
2. ${\tan }^{-1}\sqrt{3}+{\tan }^{-1}1=-{\tan }^{-1}(2+\sqrt{3})$
Which of the above is/are correct?

The correct option is C Both 1 and 2
Solution:

$1.{\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}(\frac{4}{5}\times \frac{4}{5}+\frac{3}{5}\times \frac{3}{5})$

$\because {\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$

$={\sin }^{-1}(\frac{16}{25}+\frac{9}{25})$

$={\sin }^{-1}1=\frac{\pi }{2}$

$2.{\tan }^{-1}\sqrt{3}+{\tan }^{-1}1={\tan }^{-1}\frac{\sqrt{3}+1}{1-\sqrt{3}}$

$=-{\tan }^{-1}\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=-{\tan }^{-1}(2+\sqrt{3})$ ..... [By rationalization]

Both 1 and 2 are correct.

Hence, C is the correct option.