Chapter 5.2, Problem 45E

Solution

To prove: The identity $\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=1+\sec x\csc x$ .

Given information:

The given identityis $\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=1+\sec x\csc x$ .

Formula used:

The trigonometric identities are ${\sin }^{2}x+{\cos }^{2}x=1$ , $\tan x=\frac{\sin x}{\cos x}$ , $\cot x=\frac{\cos x}{\sin x}$ , $\csc x=\frac{1}{\sin x}$ , and $\sec x=\frac{1}{\cos x}$ ,.

The algebraic identity $\left(a-b\right)\left(a^2+b^2+ab\right)=a^3-b^3$ .

Proof:

Rewrite the left-hand side of identity in terms of sine and cosine functions andsimplify using the trigonometric and algebraic identities.

  $\begin{array}{c}\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=\frac{\frac{\sin x}{\cos x}}{1-\frac{\cos x}{\sin x}}+\frac{\frac{\cos x}{\sin x}}{1-\frac{\sin x}{\cos x}}\\ =\frac{{\sin }^{2}x}{\cos x\left(\sin x-\cos x\right)}+\frac{{\cos }^{2}x}{\sin x\left(\cos x-\sin x\right)}\\ =\frac{{\sin }^{3}x-{\cos }^{3}x}{\sin x\cos x\left(\sin x-\cos x\right)}\\ =\frac{\left(\sin x-\cos x\right)\left({\sin }^{2}x+{\cos }^{2}x+\sin x\cos x\right)}{\sin x\cos x\left(\sin x-\cos x\right)}\end{array}$

Again, simplify using the trigonometric identities.

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

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

Hence, it is proved that $\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=1+\sec x\csc x$ .