The maximum value.

Phew!! Did it finally!!:):)

Okay, so for this expression to attain a maximum value, the expression in the denominator must have the minimum value. Now, for an expression to attain the minimum value, it's derivative must be equal to 0. Differentiating with respect to $\theta$ ,

$\frac { d }{ d\theta } sin^{ 2 }{ \theta }+3\sin { \theta } \cos { \theta } +5cos^{ 2 }{ \theta }=0\\ \Rightarrow \frac { d }{ d\theta } sin^{ 2 }{ \theta }+3\frac { d }{ d\theta } \sin { \theta } \cos { \theta } +5\frac { d }{ d\theta } cos^{ 2 }{ \theta }=0\\ \Rightarrow 2\sin { \theta } \frac { d }{ d\theta } \sin { \theta } +3\{ \sin { \theta } \frac { d }{ d\theta } \cos { \theta } +\cos { \theta } \frac { d }{ d\theta } \sin { \theta } \} +10\cos { \theta } \frac { d }{ d\theta } \cos { \theta } =0\\ \Rightarrow 2\sin { \theta } \cos { \theta } +3\{ cos^{ 2 }{ \theta }-sin^{ 2 }{ \theta }\} -10\sin { \theta } \cos { \theta } =0\\ \Rightarrow 3\{ cos^{ 2 }{ \theta }-sin^{ 2 }{ \theta }\} =8\sin { \theta } \cos { \theta } \quad <------1\\$

$Also,\quad cos^{ 2 }{ \theta }+sin^{ 2 }{ \theta }=1\\ sin^{ 2 }{ \theta }=1-cos^{ 2 }{ \theta }\quad <-----2$

Using 2 and 1,

$3\{ cos^{ 2 }{ \theta }-1+cos^{ 2 }{ \theta }\} =8\sin { \theta } \cos { \theta } \\ 3\{ 2cos^{ 2 }{ \theta }-1\} =4\{ 2\sin { \theta } \cos { \theta } \} \\ 3\cos { 2\theta } =4\sin { 2\theta } \\ \tan { 2\theta } =\frac { 3 }{ 4 } \\ \frac { 2\tan { \theta } }{ 1-\tan ^{ 2 }{ \theta } } =\frac { 3 }{ 4 } \\ 8\tan { \theta } =3-3\tan ^{ 2 }{ \theta } \\ 3\tan ^{ 2 }{ \theta } +8\tan { \theta } -3=0\\ Solving\quad this\quad quadratic\quad eq.,\\ \tan { \theta } =-3,\frac { 1 }{ 3 } \\$

Now, if $\tan { \theta } =\frac { 1 }{ 3 }$ , then $\theta$ lies in First quadrant. Therefore both $\sin { \theta }$ & $\cos { \theta }$ are positive and as a result the value of the expression increases. But, if $\tan { \theta } =-3$ , then $\theta$ lies in Second quadrant and here, $\cos { \theta }$ is negative whereas $\sin { \theta }$ is positive. As a result the value of this expression decreases and the value of $\frac { 1 }{ sin ^{ 2 }{ \theta } +3\sin { \theta } \cos { \theta } +5cos ^{ 2 }{ \theta } }$ increases.

So, that is for sure that we need to take $\tan { \theta } =-3$ as our value. Making a triangle with these values, $\sin { \theta } =\frac { 3 }{ \sqrt { 10 } }$ and $\cos { \theta } =\frac { -1 }{ \sqrt { 10 } }$ . Putting these values into $\frac { 1 }{ sin ^{ 2 }{ \theta } +3\sin { \theta } \cos { \theta } +5cos ^{ 2 }{ \theta } }$ , the value comes out to be 2.

Cheers!!!:):)