Question

if y=2/sin theta+square root of 3 cos theta,then the minimum value of y is​

Answer 1

Answer:

The value of y = 1

Step-by-step explanation:

The value of y which is a division term will be minimum only when the denominator is maximum to the numerator,

∴ $sin \theta +\sqrt { 3 cos\theta }$ 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 get:

$1-\sqrt { 3tan\theta } =0$

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

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

$\Rightarrow\theta=30^0$

Putting this value to the value of y, we get :

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

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

$\Rightarrow\quad y =1$