Answer
13 (a). $ 2 \sqrt 2 - 3$
13 (b). $(\sec x - \tan x)^{2}$
Work Step by Step
13 (a). Given expression is-
$\frac{1 - \sqrt 2}{1 + \sqrt 2}$
= $\frac{1 - \sqrt 2}{1 + \sqrt 2}. \frac{1 - \sqrt 2}{1 - \sqrt 2}$
{Multiplying the numerator and denominator of the fraction by the conjugate of the denominator, $(1 - \sqrt 2)$}
= $\frac{(1 - \sqrt 2)^{2}}{(1 - \sqrt 2)(1 +\sqrt 2)}$
= $\frac{1 - 2\sqrt 2 + 2}{{1^{2} - (\sqrt 2)^{2}}}$
{Recall $(a-b)^{2} = a^{2} -2ab +b^{2}$ and
$(a-b)(a+b) $ = $a^{2} - b^{2}$ }
= $\frac{ 3 - 2 \sqrt 2}{1 - 2}$
= $\frac{3 - 2 \sqrt 2}{-1}$
= $- (3 - 2 \sqrt 2)$
= $ 2 \sqrt 2 - 3$
13 (b). Given expression is-
$\frac{1 -\sin x}{1 + \sin x}$
= $\frac{1 -\sin x}{1 + \sin x}. \frac{1 -\sin x}{1 -\sin x}$
{Multiplying the numerator and denominator of the fraction by the conjugate of the denominator, $(1 -\sin x)$}
= $\frac{(1 - \sin x)^{2}}{(1 + \sin x)(1 -\sin x)}$
= $\frac{1 - 2\sin x + \sin^{2}x}{{1^{2} - (\sin x)^{2}}}$
{Recall $(a-b)^{2} = a^{2} -2ab +b^{2}$ and
$(a-b)(a+b) $ = $a^{2} - b^{2}$ }
= $\frac{1 - 2\sin x + \sin^{2}x}{1 - \sin ^{2} x}$
= $\frac{1 - 2\sin x + \sin^{2}x}{\cos^{2} x}$
( From first Pythagorean identity)
= $\frac{1}{\cos^{2} x} - \frac{2 \sin x}{\cos^{2} x} + \frac{\sin^{2} x}{\cos^{2} x}$
= $\sec^{2} x - 2.\frac{1}{\cos x}.\frac{\sin x}{\cos x} + \tan^{2} x $
(using ratio and reciprocal identities)
= $\sec^{2} x - 2.\sec x \tan x + \tan^{2} x $
= $(\sec x - \tan x)^{2}$