Chapter 7, Problem 8RE

To determine

To prove: The identity $\left(1-\tan x\right)\left(1-\cot x\right)=2-\sec x\csc x$ .

Explanation of Solution

Given information:

The given identity is $\left(1-\tan x\right)\left(1-\cot x\right)=2-\sec x\csc x$ .

Formula used:

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

Reciprocal identity:

  $\begin{array}{l}\sec x=\frac{1}{\cos x}\\ \csc x=\frac{1}{\sin x}\end{array}$

Proof:

To prove the identity, take left side of the identity and simplify it.

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

Further simplify the equation.

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

From the above, it is clear that the left hand side of the identity is equal to the right side of the identity.

Hence, proved.