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