To prove:
[tex]$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1[/tex]
Solution:
[tex]$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}[/tex]
Multiply first term by [tex]\frac{1+cos x}{1+cos x}[/tex] and second term by [tex]\frac{1-cos x}{1-cos x}[/tex].
[tex]$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}[/tex]
Using the identity: [tex](a-b)(a+b)=(a^2-b^2)[/tex]
[tex]$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}[/tex]
Denominators are same, you can subtract the fractions.
[tex]$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}[/tex]
Using the identity: [tex]1-\cos ^{2}(x)=\sin ^{2}(x)[/tex]
[tex]$= \frac{1+\cos^2 x}{\sin^2x}[/tex]
Using the identity: [tex]1=\cos ^{2}(x)+\sin ^{2}(x)[/tex]
[tex]$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}[/tex]
[tex]$=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x}[/tex] ------------ (1)
[tex]RHS=2 \cot ^{2} x+1[/tex]
Using the identity: [tex]\cot (x)=\frac{\cos (x)}{\sin (x)}[/tex]
[tex]$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}[/tex]
[tex]$=1+2\frac{\cos^{2} x}{\sin^{2} x}[/tex]
[tex]$=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x}[/tex] ------------ (2)
Equation (1) = Equation (2)
LHS = RHS
[tex]$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1[/tex]
Hence proved.
