Question

Answer this Guyz with soln...Brainliest For The Best​

Answer 1

Solution:

$f( \theta) = \frac{1}{ { \sin }^{2} \theta + 3 \sin \theta \cos \theta + 5 \cos ^{2} \theta}$

$= \frac{1}{( { \sin }^{2} \theta + \cos ^{2} \theta)+ 2({( 2\cos}^{2} \theta - 1) + 1) + 3 \sin \theta \cos \theta}$

$= \frac{1}{1 + 2( \cos(2 \theta) + 1) + \frac{3}{2} \sin(2 \theta) }$

$= \frac{1}{ \frac{3}{2} \sin(2 \theta) + 2 \cos(2 \theta) + 3}$

Since;

$c - \sqrt{a^{2} + {b}^{2} } \leq a (\sin(x) + b \cos(x) + c) \leq c + \sqrt{ {a}^{2} + {b}^{2} }$

$3 - \sqrt{( \frac{3}{2} ) ^{2} + (2) ^{2} } \leq\frac{1}{f( \theta)} \leq 3 + \sqrt{( \frac{3}{2}) ^{2} + (2) ^{2} }$

$\frac{1}{3 - \frac{5}{2} } \geq f( \theta) \geq \frac{1}{3 + \frac{5}{2} }$

$2 \geqslant f( \theta) \geqslant \frac{2}{11}$

OR

$f( \theta)∈ [ \frac{2}{11},2 ]$

∴ ,

$f( \theta) | _{min} = \frac{2}{11} \: , and \: f( \theta)| _{max} = 2$

Answer 2

Ryan oppa ಥ_ಥ

Do u remember me???????