Question

Trig proofs with Pythagorean Identities.

Answer 1

To prove:

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$.

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

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

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

Denominators are same, you can subtract the fractions.

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

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

       $= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

       $=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

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

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

        $=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

       $=1+2\frac{\cos^{2} x}{\sin^{2} x}$

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

Equation (1) = Equation (2)

LHS = RHS

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.