Answer :
Answer:
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.
Explanation:
Given the below identity;
[tex]\mleft(\cot x\mright)\mleft(\cos x\mright)=cscx-\sin x[/tex]Let's analyze the steps Student A used in proving the above identity;
Step 1:
[tex]\frac{\cos x}{\sin x}\cdot\cos x=\csc x-\sin x[/tex]Recall;
[tex]\begin{gathered} \tan x=\frac{\sin x}{\cos x}\text{ and cot x }=\frac{1}{\tan x} \\ \therefore\text{cot x }=\frac{1}{\frac{\sin x}{\cos x}}=1\div\frac{\sin x}{\cos x}=1\cdot\frac{\cos x}{\sin x}=\frac{\cos x}{\sin x} \end{gathered}[/tex]Step 2:
[tex]\frac{\cos^2x}{\sin x}=\csc x-\sin x[/tex]Step 3:
[tex]\frac{1-\sin ^2x}{\sin x}=\csc x-\sin x[/tex]The below trig identity was applied here;
[tex]\begin{gathered} \cos ^2x+\sin ^2x=1 \\ \therefore\cos ^2x=1-\sin ^2x \end{gathered}[/tex]Step 4:
[tex]\frac{1}{\sin x}-\frac{\sin ^2x}{\sin x}=\csc x-\sin x[/tex]Step 5:
[tex]\csc x-\sin x=\csc x-\sin x[/tex]Note that;
[tex]\begin{gathered} \frac{1}{\sin x}=\csc x \\ \text{and} \\ \frac{\sin^2x}{\sin x}=\sin x \end{gathered}[/tex]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;
[tex]\csc x=\frac{1}{\sin x}[/tex]So Step 1 is given as;
[tex]\cot x\cos x=\frac{1}{\sin x}-\sin x[/tex]Step 2:
Note that;
[tex]\frac{\sin^2x}{\sin x}=\sin x[/tex]So Step 2 is given as;
[tex]\cot x\cos x=\frac{1}{\sin x}-\frac{\sin^2x}{\sin x}[/tex]Step 3:
Using sin x which is the LCM to multiply through, we'll have;
[tex]\cot x\cos x=\frac{1-\sin^2x}{\sin x}[/tex]Step 4:
Recall the below trig identity;
[tex]\begin{gathered} \cos ^2x+\sin ^2x=1 \\ \therefore\cos ^2x=1-\sin ^2x \end{gathered}[/tex]Applying the above identity, we'll have;
[tex]\cot x\cos x=\frac{\cos^2x}{\sin x}[/tex]Step 5:
Note that;
[tex]\cos ^2x=\cos x\cdot\cos x[/tex]So we'll have;
[tex]\cot x\cos x=\frac{\cos x}{\sin x}\cdot\cos x[/tex]Step 6;
Recall that;
[tex]\begin{gathered} \tan x=\frac{\sin x}{\cos x} \\ \text{and} \\ \cot x=\frac{1}{\tan x}=\frac{1}{\frac{\sin x}{\cos x}}=1\div\frac{\sin x}{\cos x}=1\cdot\frac{\cos x}{\sin x}=\frac{\cos x}{\sin x} \\ \therefore\cot x=\frac{\cos x}{\sin x} \end{gathered}[/tex]So we'll have;
[tex]\cot x\cos x=\cot x\cos x[/tex]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.