Question

plz solve this fast​

Answer 1

$- 2sin \frac{x}{2} = \sqrt{1 + sin \: x } + \sqrt{1 - sin \: x} \\ \sqrt{1 + sinx} = \sqrt{sin {}^{2} \frac{x}{2} + cos {}^{2} \frac{x}{2} + 2sin \frac{x}{2}cos \frac{x}{2} } \\ = \sqrt{ (sin \frac{x}{2} + cos \frac{x}{2} ) {}^{2} } \\ = (sin \frac{x}{2} + cos \frac{x}{2} ) \\ \sqrt{1 - sinx} = \sqrt{sin {}^{2} \frac{x}{2} + cos {}^{2} \frac{x}{2} - 2sin \frac{x}{2}cos \frac{x}{2} } \\ = \sqrt{ (sin \frac{x}{2} - cos \frac{x}{2} ) {}^{2} } \\ = (sin \frac{x}{2} - cos \frac{x}{2} ) \\ adding \: \\ = 2sin \frac{x}{2} \\ as \: x = 460 \\ \frac{x}{2} = 230(iii \: qudrnt) \\ sin \: is \: - ve \: so \\ = - 2sin \frac{x}{2}$