Question

If y is equal to 2 by sin theta + root 3 cos theta then the minimum value of y is

Answer 1

Answer:

Given

$y = \frac{2}{sin \alpha + \sqrt{3} \: cos \alpha }$

we know that

$- \sqrt{ {1}^{2} + { \sqrt{3} }^{2} } \leqslant sin \alpha + \sqrt{3} \: cos \alpha \leqslant \sqrt{ {1}^{2} + { \sqrt{3} }^{2} }$

$- 2 \leqslant sin \alpha + \sqrt{3} \: cos \alpha \leqslant 2$

$- \frac{1}{2} \geqslant \frac{1}{sin \alpha + \sqrt{3} \: cos \alpha } \geqslant \frac{1}{2}$

$- 1\geqslant \frac{2}{sin \alpha + \sqrt{3} \: cos \alpha } \geqslant1$

so minimum Value & maximum Value not define.