Question

Prove that sin30 + cos60 by 1+ sin60 + cos30 = 1 by 2 ( root 3 - 1)

Answer 1

Step-by-step explanation:

$\frac{sin \: 30 \degree + cos \: 60 \degree}{1 + sin \: 60 \degree + cos \: 30 \degree} = \frac{1}{2}( \sqrt{3} - 1) \\ \\ LHS = \frac{sin \: 30 \degree + cos \: 60 \degree}{1 + sin \: 60 \degree + cos \: 30 \degree} \\\\ \huge=\frac{ \frac{1}{2} + \frac{1}{2} }{1 + \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{2} } \\ \\ \huge=\frac{ \frac{1 + 1}{2} }{1 + \frac{ \sqrt{3} + \sqrt{3} }{2} } \\ \\ \huge=\frac{ \frac{2}{2} }{1 + \frac{2 \sqrt{3} }{2} } \\ \\ \huge= \frac{1}{1 + \sqrt{3} } \\ \\ \huge= \frac{1}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} \\\\ \huge = \frac{1 - \sqrt{3} }{ {1}^{2} - { \sqrt{3} }^{2} } \\ \\ \huge = \frac{1 - \sqrt{3} }{ 1- 3 } \\\\ \huge = \frac{1 - \sqrt{3} }{ - 2 } \\\\\huge= \frac{ - ( \sqrt{3} - 1)}{ - 2}\\ \\ \huge= \frac{ \sqrt{3} - 1 }{2} \\ \\ \huge= \frac{1}{2} ( \sqrt{3} - 1) \\\\ \huge = RHS \\\\ \huge{ thus \: proved}$