Rewriting trigonometric expression in terms of $\cot x$
Rewrite the following expression in terms of $\cot x$:
$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$$
I usually show my work on this site but I'm really lost about this problem. Any help would be greatly appreciated. Thanks.
$\endgroup$ 34 Answers
$\begingroup$$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} = \frac{1+\cos x-\cos x +\cos^2 x}{1-\cos^2 x} = \frac{1+\cos^2 x}{\sin^2 x} = \frac{\sin^2 x + 2\cos^2 x}{\sin^2 x} =\\ =1 + 2\cot^2 x$$
$\endgroup$ $\begingroup$$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=\frac{1+\cos x}{1-\cos^2 x}-\frac{\cos x- \cos^2 x}{1-\cos^2 x}=\frac{1+\cos^2 x}{1-\cos^2 x}=\frac{1+\cos^2 x}{\sin^2 x}=\frac{2\cos^2 x+ \sin^2}{\sin^2 x}=2\cot^2 x +1$$
First you get common denominator $1-\cos^2 x=(1+\cos x)(1-\cos x)$, in next steps you use $\cos^2 x+ \sin^2 x=1$.
$\endgroup$ $\begingroup$$$\require{cancel}\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} \\ \frac{1}{1-\cos x}\color{red}{\frac{1+\cos x}{1+\cos x}}-\frac{\cos x}{1+\cos x}\color{red}{\frac{1-\cos x}{1-\cos x}}\\\frac{1+\cos x}{1-\cos^2x}-\frac{\cos x-\cos^2 x}{1-\cos ^2x}\\\frac{1+\cancel{\cos x-\cos x}+\cos^2x}{1-\cos^2x} \\ \frac{1+\cos^2x}{\sin^2x}\\\therefore \csc^2x+\cot^2x$$
If only $\cot^2x$:
$$\frac{1+\cos^2x}{\sin^2x}\\ \frac{\sin^2x+\cos^2x+\cos^2x}{\sin^2x}\\\therefore 1+2\cot^2 x$$
Or you could have just used another identity.
$\endgroup$ 2 $\begingroup$First convert everything to $\csc$s and $\cot$s by dividing numerators and denominators by $1/\sin x$. $$\begin{array}{lll} \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}&=&\frac{1}{1-\cos x}\cdot\frac{1/\sin x}{1/\sin x}-\frac{\cos x}{1+\cos x}\cdot\frac{1/\sin x}{1/\sin x}\\ &=&\frac{1/\sin x}{1/\sin x-\cos x/\sin x}-\frac{\cos x/\sin x}{1/\sin x+\cos/\sin x}\\ &=&\frac{\csc x}{\csc x-\cot x}-\frac{\cot x}{\csc x+\cot x}\\ &=&\frac{\csc x}{\csc x-\cot x}\cdot\frac{\csc x+\cot x}{\csc x+\cot x}-\frac{\cot x}{\csc x+\cot x}\cdot\frac{\csc x-\cot x}{\csc x-\cot x}\\ &=&\frac{\csc^2 x+\csc x\cot x}{\csc^2 x-\cot^2 x}-\frac{\csc x\cot x-\cot^2 x}{\csc^2 x-\cot^2 x}\\ &=&\frac{\csc^2 x+\cot^2 x}{(\cot^2 x+1)-\cot^2 x}\\ &=&\csc^2 x+\cot^2 x\\ &=&(\cot^2 x+1)+\cot^2 x\\ &=&2\cot^2 x+1 \end{array}$$
$\endgroup$
