Trigonometry! #13

Using $\csc^{2} \theta = 1 + \cot^{2} \theta$ and $sec^{2} \theta= 1+\tan^{2} \theta$ , the expression can be simplified to $\frac {37}{24} + \frac {2}{3} \cot^{2} \theta + \frac {3}{8} \tan^{2} \theta$ .

Now, for $\frac {2}{3} \cot^{2} \theta + \frac {3}{8} \tan^{2} \theta$ , we use A.M.-G.M. inequality.

$\frac {2}{3} \cot^{2} \theta + \frac {3}{8} \tan^{2}\theta ≥ 2\sqrt{\left(\frac {2}{3} \cot^{2}\theta \right)\left(\frac {3}{8} \tan^{2} \theta \right)}$

$\frac {2}{3} \cot^{2} \theta + \frac {3}{8} \tan^{2}\theta ≥ 1$

Hence the minimum value of the expression is $\frac {37}{24} + 1 = \boxed {\frac{61}{24}}$

Let $\space f(\theta) = \frac{1}{2}+\frac{2}{3}\csc^2{\theta}+\frac{3}{8}\sec^2{\theta} = \frac{1}{2}+\frac{2}{3\sin^2{\theta}}+\frac{3}{8\cos^2{\theta}}$ .

Then, $\space \dfrac {df(\theta)}{d\theta} = \dfrac {(-2)(2 \cos{\theta})}{3\sin^3{\theta}} + \dfrac {(-2)(3)(-\sin{\theta})}{8\cos^3{\theta}} = \dfrac {9\sin^4{\theta} - 16\cos^4{\theta}}{24\sin^3{\theta}\cos^3{\theta}}$

When $\space f(\theta)$ is minimum, $\space \frac {df(\theta)}{d\theta} = 0$

$\Rightarrow 9\sin^4{\theta} - 16\cos^4{\theta} = 0\quad \Rightarrow \tan^4 {\theta} = \frac {16}{9}\quad \Rightarrow \tan^2{\theta} = \frac {4}{3}$

$\Rightarrow \sin^2{\theta} = \frac {4}{7}\quad \Rightarrow \cos^2{\theta} = \frac {3}{7} \quad \Rightarrow f(\theta) = \frac{1}{2}+\frac{2}{3}(\frac{7}{4}) +\frac{3}{8}(\frac{7}{3}) = \boxed{\frac{61}{24}}$