How do you verify $\frac{{(cos x - 1)}^{2}}{{sin}^{2} x} = \frac{1 - cos x}{1 + cos x}$ $[(cosx - 1)^2] / [sin^2 x] = [1 - cos x] / [1 + cos x]$ ?

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Answer 1 · 2015-08-02T17:17:56

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

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

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

How do you verify (cosx−1)2sin2x=1−cosx1+cosx[(cosx - 1)^2] / [sin^2 x] = [1 - cos x] / [1 + cos x]?