Question

If y=2/sin thita + root 3 cos thita then minimum value of y is

Answer 1

i have answered it and it is in the clicked picture which i have posted

Answer 2

The value of y which is a division term will be minimum only when the denominator is maximum to the numerator, therefore $sin \theta +\sqrt { 3 cos\theta }$ needs to be maximum which will in case

$\frac { d }{ d\theta } (sin\theta +\sqrt { 3cos\theta } )=0\Rightarrow\frac { d }{ d\theta } (sin\theta )+\frac { d }{ d\theta } (\sqrt { 3cos\theta } )=0$

$\Rightarrow\frac { d }{ d\theta } (sin \theta )+\sqrt { 3\frac { d }{ d\theta } (cos\theta ) }$

$\Rightarrow\quad cos \theta -\sqrt { 3sin\theta } =0$

Divide the equation by $cos\theta , we\quad get\quad 1-\sqrt { 3tan\theta } =0$

$\Rightarrow\quad -\sqrt { 3 tan \theta } =-1$

$\Rightarrow\quad tan \theta \quad =\quad \frac { 1 }{ \sqrt { 3 } } \quad$

$\Rightarrow\theta=30^0.$

Putting this value to the value of y, we get

$-y=\frac { 2 }{ sin30^0+\sqrt { 3 } cos30^0 }$

$\Rightarrow\quad y=\frac { 2 }{ [\frac { 1 }{ 2 } +\sqrt { 3x } (\sqrt { \frac { 3 }{ 2 } } )] }$

$\Rightarrow\quad y =1.$