Mathematics - Trigonometric Equation Question with Solution | TestHub

Given, $8^{2{\sin }^2\theta }+8^{2{\cos }^2\theta }=16$

$\Rightarrow 8^{2{\sin }^2\theta }+8^{2-2{\sin }^2\theta }=16$

Now let $8^{2{\sin }^2\theta }=y$ $$

$\Rightarrow y+\frac{64}{y}=16$

$\Rightarrow y=8$

$\Rightarrow 8^{2{\sin }^2\theta }=8$

$\Rightarrow {\sin }^2\theta =\frac{1}{2}$

$\Rightarrow \theta \in \left\{\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right\}$

Now $n\left(S\right)+\sum _{\theta \in \mathrm{S}}\left(\sec \left(\frac{\pi }{4}+2\theta \right)cosec\left(\frac{\pi }{4}+2\theta \right)\right)$

 $=\mathrm{n}\left(\mathrm{S}\right)+\sum _{\theta \in \mathrm{S}}\frac{1}{\cos \left({\displaystyle \frac{\pi }{4}}+2\theta \right)\sin \left({\displaystyle \frac{\pi }{4}}+2\theta \right)}$

$=4+\sum _{\theta \in \mathrm{S}}\frac{2}{2\cos \left({\displaystyle \frac{\pi }{4}+2\theta }\right)\sin \left({\displaystyle \frac{\pi }{4}+2\theta }\right)}$

$=4+\sum _{\theta \in \mathrm{S}}2\cos \text{ec}\left(\frac{\pi }{2}+4\theta \right)$

$=4+2\cos \text{ec}\left(\frac{\pi }{2}+\mathrm{\pi }\right)\text{+}2\cos \text{ec}\left(\frac{\pi }{2}+3\mathrm{\pi }\right)\text{+}2\cos \text{ec}\left(\frac{\pi }{2}+5\mathrm{\pi }\right)\text{+}2\cos \text{ec}\left(\frac{\pi }{2}+7\mathrm{\pi }\right)$

$=4+2\left[-\cos \text{ec}\frac{\mathrm{\pi }}{2}-\cos \text{ec}\frac{\mathrm{\pi }}{2}-\cos \text{ec}\frac{\mathrm{\pi }}{2}-\cos \text{ec}\frac{\mathrm{\pi }}{2}\right]$

$=4+\left(-2\right)\times 4=-4$