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