Let f(x) = 25 ^ x 25 ^ x + 5 , then the number of solution(s) of...

$f\left(\sin^{2}\theta\right) + f$\left(1 - \sin^{2}\theta\right)
Let, $\sin^{2}\theta = t$
$∴ f\left(t\right)$ + f\left(1 - t\right)$<br />$= \dfrac{{25}^{t}}{{25}^{t} + 5} + \dfrac{{25}^{1 - t}}{{25}^{1 - t} + 5} = \dfrac{{25}^{t}}{{25}^{t} + 5} + \dfrac{25}{25 + 5\left({25}^{t}\right)}$<br />$= \dfrac{{25}^{t} + 5}{{25}^{t} + 5} = 1$<br />Therefore, the given equation will become$\tan^{2}\theta = 1$<br />$∴ tan \theta = \pm 1 ⇒ \theta = n\pi \pm \dfrac{\pi}{4}, n \in Z$<br />Hence, the number of solutions are<br />$4 \times 5 = 20$<br />{∵ 4 solutions in \left[0,2\pi\right]}$