Trigonometri #1

Lets start with the expression that we have to solve. $\dfrac{1}{\cos^{2}\theta} +\dfrac{1}{1+\sin^{2}\theta} + \dfrac{2}{1+\sin^{4}\theta}+\dfrac{4}{1+\sin^{8}\theta}$ Put $\cos^{2}\theta = 1-\sin^{2}\theta$ . $\dfrac{1}{1-\sin^{2}\theta} + \dfrac{1}{1+\sin^{2}\theta} + \dfrac{2}{1+\sin^{4}\theta}+\dfrac{4}{1+\sin^{8}\theta}$ Combing first two terms:- $\dfrac{2}{1-\sin^{4}\theta}+\dfrac{2}{1+\sin^{4}\theta}+\dfrac{4}{1+\sin^{8}\theta}$ Again combing first two terms:- $\dfrac{4}{1-\sin^{8}\theta}+\dfrac{4}{1+\sin^{8}\theta}$ Again combing these two terms:- $\dfrac{8}{1-\sin^{16}\theta}$ Now we know the value of $\sin^{16}\theta = \dfrac{1}{5}$ . Replacing it in above expression:- $\dfrac{8}{1-\dfrac{1}{5}}$ Multiplying numerator and denominator by $5$ :- $=\dfrac{40}{5-1}$ $=\dfrac{40}{4}$ $=\boxed{10}$

Simply replace (costheta)^{2} by (1-sintheta)^{2} and solve from left to right