Question

Students were asked to prove the identity (cot x)(cos x) = csc x − sin x. Two students' work is given.Did either student verify the identity properly? Explain why or why not.

Answer 1

Yes, both students A and B verified the identity properly.

Because in the final step, the expressions on both sides of the equation were equal ( RHS = LHS) and the correct trig identities were utilized in the course of the solution.

Step-by-step explanation:

Given the below identity;

`\mleft(\cot x\mright)\mleft(\cos x\mright)=cscx-\sin x`

Let's analyze the steps Student A used in proving the above identity;

Step 1:

`(\cos x)/(\sin x)\cdot\cos x=\csc x-\sin x`

Recall;

`\begin{gathered} \tan x=(\sin x)/(\cos x)\text{ and cot x }=(1)/(\tan x) \\ \therefore\text{cot x }=(1)/((\sin x)/(\cos x))=1/(\sin x)/(\cos x)=1\cdot(\cos x)/(\sin x)=(\cos x)/(\sin x) \end{gathered}`

Step 2:

`(\cos^2x)/(\sin x)=\csc x-\sin x`

Step 3:

`(1-\sin ^2x)/(\sin x)=\csc x-\sin x`

The below trig identity was applied here;

`\begin{gathered} \cos ^2x+\sin ^2x=1 \\ \therefore\cos ^2x=1-\sin ^2x \end{gathered}`

Step 4:

`(1)/(\sin x)-(\sin ^2x)/(\sin x)=\csc x-\sin x`

Step 5:

`\csc x-\sin x=\csc x-\sin x`

Note that;

`\begin{gathered} (1)/(\sin x)=\csc x \\ \text{and} \\ (\sin^2x)/(\sin x)=\sin x \end{gathered}`

We can see from the above that Student A verified the identity properly because in the final step, the expressions on both sides of the equation were equal ( RHS = LHS) and the correct trig identities were utilized.

Let's analyze the steps of Student V;

Step 1:

Recall that;

`\csc x=(1)/(\sin x)`

So Step 1 is given as;

`\cot x\cos x=(1)/(\sin x)-\sin x`

Step 2:

Note that;

`(\sin^2x)/(\sin x)=\sin x`

So Step 2 is given as;

`\cot x\cos x=(1)/(\sin x)-(\sin^2x)/(\sin x)`

Step 3:

Using sin x which is the LCM to multiply through, we'll have;

`\cot x\cos x=(1-\sin^2x)/(\sin x)`

Step 4:

Recall the below trig identity;

`\begin{gathered} \cos ^2x+\sin ^2x=1 \\ \therefore\cos ^2x=1-\sin ^2x \end{gathered}`

Applying the above identity, we'll have;

`\cot x\cos x=(\cos^2x)/(\sin x)`

Step 5:

Note that;

`\cos ^2x=\cos x\cdot\cos x`

So we'll have;

`\cot x\cos x=(\cos x)/(\sin x)\cdot\cos x`

Step 6;

Recall that;

`\begin{gathered} \tan x=(\sin x)/(\cos x) \\ \text{and} \\ \cot x=(1)/(\tan x)=(1)/((\sin x)/(\cos x))=1/(\sin x)/(\cos x)=1\cdot(\cos x)/(\sin x)=(\cos x)/(\sin x) \\ \therefore\cot x=(\cos x)/(\sin x) \end{gathered}`

So we'll have;

`\cot x\cos x=\cot x\cos x`

We can see from the above that Student B verified the identity properly because in the final step, the expressions on both sides of the equation were equal ( RHS = LHS) and the correct trig identities were utilized.