Solution:
The question is given below as
![cotx(cosx)=cscx-sinx](https://img.qammunity.org/2023/formulas/mathematics/college/k8qswpjuemj1pcphp3ifu6ug2wdyb8hdvs.png)
Step 1:
We will make use of the quotient identity below
![cotx=(cosx)/(sinx)](https://img.qammunity.org/2023/formulas/mathematics/college/ehwl5edttypnvd2jkefjps8bh8gnz08ik9.png)
![\begin{gathered} cotx(cosx)=cscx-s\imaginaryI nx \\ (cosx)/(sinx)(cosx)=cscx-s\mathrm{i}nx \\ (cos^2x)/(sinx)=cscx-s\mathrm{i}nx \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r79ihvq8ot7xiww2cadrge988coqlvtby9.png)
Step 2:
We will make use of the Pythagorean identity below
![\begin{gathered} cos^2x+sin^2x=1 \\ cos^2x=1-sin^2x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kes0onrybxjgxkuuqfv5gj8t740guhxbsf.png)
![\begin{gathered} (cos^(2)x)/(s\imaginaryI nx)=cscx-s\imaginaryI nx \\ (1-sin^2x)/(sinx)=cscx-s\mathrm{i}nx \\ (1)/(sinx)-(sin^2x)/(sinx)=cscx-s\mathrm{i}nx \\ cscx-sinx=cscx-s\mathrm{i}nx(PROVED) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yzgzli0nni2o3bkzc2kliwb8oudu3qb6il.png)
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
![cotx=(cosx)/(sinx)](https://img.qammunity.org/2023/formulas/mathematics/college/ehwl5edttypnvd2jkefjps8bh8gnz08ik9.png)
Pythagorean identity was used in step 3
![\begin{gathered} cos^2x+sin^2x=1 \\ cos^2x=1-sin^2x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kes0onrybxjgxkuuqfv5gj8t740guhxbsf.png)
Reciprocal identity was used in step 5
![cscx=(1)/(sinx)](https://img.qammunity.org/2023/formulas/mathematics/college/464sgpbohn7b8l6jsi8upusg9ctw9gtufm.png)