Solution:
The question is given below as
[tex]cotx(cosx)=cscx-sinx[/tex]
Step 1:
We will make use of the quotient identity below
[tex]cotx=\frac{cosx}{sinx}[/tex][tex]\begin{gathered} cotx(cosx)=cscx-s\imaginaryI nx \\ \frac{cosx}{sinx}(cosx)=cscx-s\mathrm{i}nx \\ \frac{cos^2x}{sinx}=cscx-s\mathrm{i}nx \end{gathered}[/tex]
Step 2:
We will make use of the Pythagorean identity below
[tex]\begin{gathered} cos^2x+sin^2x=1 \\ cos^2x=1-sin^2x \end{gathered}[/tex][tex]\begin{gathered} \frac{cos^{2}x}{s\imaginaryI nx}=cscx-s\imaginaryI nx \\ \frac{1-sin^2x}{sinx}=cscx-s\mathrm{i}nx \\ \frac{1}{sinx}-\frac{sin^2x}{sinx}=cscx-s\mathrm{i}nx \\ cscx-sinx=cscx-s\mathrm{i}nx(PROVED) \end{gathered}[/tex]
Hence,
STUDENT A and STUDENT B both proved it properly,
STUDENT A proved it from left to right
STUDENT B proved it from right to left
For STUDENT A's work,
Quotient identity was used in step 1
[tex]cotx=\frac{cosx}{sinx}[/tex]
Pythagorean identity was used in step 3
[tex]\begin{gathered} cos^2x+sin^2x=1 \\ cos^2x=1-sin^2x \end{gathered}[/tex]
Reciprocal identity was used in step 5
[tex]cscx=\frac{1}{sinx}[/tex]
