Verify this identity: $\sin x/(1 - \cos x) = \csc x + \cot x$
Verify this identity $\frac{\sin {x}} {1 - \cos {x}}\ = \csc x + \cot x$ I got the left side to $\frac{1-\cos x} {\csc x}$, but I can't get any farther. Am I on the right path? Can I get some help?
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$\begingroup$The right-hand side can be rewritten as
$$\csc{x} + \cot{x} = \frac{1}{\sin{x}} + \frac{\cos{x}}{\sin{x}} = \frac{1 + \cos{x}}{\sin{x}}$$
Multiplying the left side top and bottom by $1 + \cos{x}$, we find that
$$\frac{\sin{x}}{1 - \cos{x}} = \frac{\sin{x}(1 + \cos{x})}{1 - \cos^2{x}} = \frac{\sin{x}(1 + \cos{x})}{\sin^2{x}} = \frac{1 + \cos{x}}{\sin{x}}$$
$\endgroup$ $\begingroup$As $\displaystyle\sin^2x+\cos^2x=1,$
$\implies \displaystyle\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$
$\implies \displaystyle\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}=\frac1{\sin x}+\frac{\cos x}{\sin x}=\csc x+\cot x$
$\endgroup$ 1 $\begingroup$To prove sin x/1-cos x = cosec x + cot x We know that, Sin^2 x = 1- cos^2 So from RHS we get, Cosec x+cot x =1/sin x + cos x/sin x So taking LCM, We get, 1+cos x/sin x Now multiply both numerator and denominator by sin x to get denominator as sin^2 x We get, (1+cos x)(sin x)/sin^2 x Now we can write sin^2 x as 1- cos^2 x So, we get (1+cos x)(sin x)/1-cos^2 x (1+cos x)(sin x)/(1+cos x)(1-cos x) applying a^2-b^2=(a+b)(a-b) so by cancellation We get sin x/1-cosx=cosec x+cot x
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