Chapter 5, Problem 1RE

To determine

To prove: The expression $\sec x-\cos x=\tan x\sin x$.

Explanation of Solution

Given information:

$\sec x-\cos x=\tan x\sin x$

Formula used:

$\sec x=\frac{1}{\cos x}$

${\sin }^2\theta +{\cos }^2\theta =1$

$\tan x=\frac{\sin x}{\cos x}$

Proof:

We have the expression in the left side $\sec x-\cos x$ can be simplified by using the reciprocal identity $\sec x=\frac{1}{\cos x}$.

$\sec x-\cos x=\frac{1}{\cos x}-\cos x$

And the expression can be further simplified by multiplying and dividing the expression by $\cos x$.

$\begin{array}{c}\sec x-\cos x=\frac{1}{\cos x}-\cos x\\ =\frac{1}{\cos x}-\frac{\cos x}{1}.\frac{\cos x}{\cos x}\\ =\frac{1-{\cos }^{2}x}{\cos x}\end{array}$

As we know the Pythagorean theory ${\sin }^2\theta +{\cos }^2\theta =1$, therefore, ${\sin }^2\theta =1-{\cos }^2\theta$. Thus, applying the theorem. And as per the quotient identity $\tan x=\frac{\sin x}{\cos x}$

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

Therefore, left side is equals to the right side.