Explanation

(a)
Given, $u_{n} = \int_{0}^{\pi/4}{\tan^{n}xdx}$
or $u_{n} = \int_{0}^{\pi/4}{\tan^{n - 2}{x\ (\sec^{2}{x - 1)\ dx}}}$
$= \int_{0}^{\pi/4}{\tan^{n - 2}{x\sec^{2}{x\ dx -}}}\int_{0}^{\pi/4}{\tan^{n - 2}{x\ dx}}$
On putting $\tan{x = t,\sec^{2}{x\ dx = dt}}$ in Ist integral, we get
$u_{n} = \int_{0}^{1}t^{n - 2}dt - u_{n - 2}$
$\Rightarrow \ \ u_{n} + u_{n - 2} = \left\lbrack \frac{t^{n - 1}}{n - 1} \right\rbrack_{0}^{1} = \left\lbrack \frac{1^{n - 1}}{n - 1} - 0 \right\rbrack = \frac{1}{n - 1}$