Chapter 6, Problem 1MCCP

In Exercises 1-18, verify each identity.

$\cos x(\tan x+\cot x)=\csc x$

To determine

To prove: The identity $\cos x\left(\tan x+\cot x\right)=\csc x$.

Explanation of Solution

Given: $\cos x\left(\tan x+\cot x\right)=\csc x$

Formula used:

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

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

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

$\frac{1}{\sin x}=\csc x$

Proof:

The expression in left side $\cos x\left(\tan x+\cot x\right)$ can be written in the form of cos and sin by using the quotient identity.

As per the quotient identity of the trignometery $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{\sin x}{\cos x}$.

Thus, expression in left side can be written and simplfied as:

$\begin{array}{ccc}\text{cos }x\left(\tan x+\cot x\right)& =& \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\hfill \\ & =& \cos x\left(\frac{\sin x}{\cos x}.\frac{\sin x}{\sin x}+\frac{\cos x}{\sin x}.\frac{\cos x}{\cos x}\right)\hfill \\ & =& \cos x\left(\frac{{\sin }^{2}x}{\sin x\cos x}+\frac{{\cos }^{2}x}{\sin x\cos x}\right)\hfill \\ & =& \cos x\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right)\hfill \end{array}$

The expression can be further calssified by applying the pythogorian identity ${\sin }^{2}x+{\cos }^{2}x=1$

$\begin{array}{ccc}\hfill \cos x\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right)& =& \cos x.\frac{1}{\sin x\cos x}\hfill \\ & =& \frac{1}{\sin x}\hfill \\ & =& \csc x\hfill \end{array}$

The expression $\frac{1}{\sin x}$ can be further simplified by applying reciprocal identity $\frac{1}{\sin x}=\csc x$.

Thus, left side is equal to right side $\cos x\left(\tan x+\cot x\right)=\csc x$.