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Answer 1 · 2017-03-03T14:43:41

$L H S = \frac{1}{1 - cos x} - \frac{cos x}{1 + cos x}$ $LHS=1 / (1- cos x) - cos x / (1 + cos x)$

$= \frac{1 + cos x - cos x (1 - cos x)}{(1 - cos x) (1 + cos x)}$ $=(1 +cos x- cos x(1-cosx)) / ((1-cosx)(1 + cos x))$

$= \frac{1 + cos x - cos x + {cos}^{2} x}{1 - {cos}^{2} x}$ $=(1 +cos x- cos x+cos^2x) / (1-cos^2x)$

$= \frac{1 + 1 - {sin}^{2} x}{{sin}^{2} x}$ $=(1 +1-sin^2x) / sin^2x$

$= \frac{2}{{sin}^{2} x} - \frac{{sin}^{2} x}{{sin}^{2} x}$ $=2/sin^2x-sin^2x/ sin^2x$

$= 2 {csc}^{2} x - 1$ $=2csc^2x-1$