Chapter 5.2, Problem 47E

Solution

To prove: The identity $\frac{2\tan x}{1-{\tan }^{2}x}+\frac{1}{2{\cos }^{2}x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}$ .

Given information:

The given identity is $\frac{2\tan x}{1-{\tan }^{2}x}+\frac{1}{2{\cos }^{2}x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}$ .

Formula used:

The trigonometric identities are ${\sin }^{2}x+{\cos }^{2}x=1$, $\sin 2x=2\sin x\cos x$, $\cos 2x={\cos }^{2}x-{\sin }^{2}x$, $\tan x=\frac{\sin x}{\cos x}$, $\tan \left(2x\right)=\frac{2\tan x}{1-{\tan }^{2}x}$ and $\cos \left(2x\right)=2{\cos }^{2}x-1$ .

The algebraic identities $\left(a-b\right)\left(a+b\right)=a^2-b^2$ and ${\left(a+b\right)}^2=a^2+b^2+2ab$ .

Proof:

Simplify the left-hand side of the given equation using the trigonometric identities as follows.

  $\begin{array}{c}\frac{2\tan x}{1-{\tan }^{2}x}+\frac{1}{2{\cos }^{2}x-1}=\tan 2x+\frac{1}{\cos 2x}\\ =\frac{\sin 2x}{\cos 2x}+\frac{1}{\cos 2x}\\ =\frac{1+\sin 2x}{\cos 2x}\\ =\frac{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}\end{array}$

Again simplify the obtained using algebraic identities as follows.

  $\begin{array}{c}\frac{2\tan x}{1-{\tan }^{2}x}+\frac{1}{2{\cos }^{2}x-1}=\frac{{\left(\sin x+\cos x\right)}^2}{\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)}\\ =\frac{\cos x+\sin x}{\cos x-\sin x}\end{array}$

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that $\frac{2\tan x}{1-{\tan }^{2}x}+\frac{1}{2{\cos }^{2}x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}$ .