Answer
The student should get 0 points out of 10 points.
Work Step by Step
Let us consider the left side of the provided expression:
$\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}$
Then, use the formula ${{A}^{2}}-{{B}^{2}}=(A+B)(A-B)$ , with $A=\sin x$ , and $B=\cos x$. The above expression can be simplified as:
$\begin{align}
& \frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}=\frac{\left( \sin x+\cos x \right)\left( \sin x-\cos x \right)}{\sin x+\cos x} \\
& =\sin x-\cos x
\end{align}$
So, the left side of the expression is equal to the right-hand side, which is
$\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}\,=\sin x-\cos x$
And the equation $\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}\,=\sin x-\cos x$ was solved by the student in the following manner:
Careful observation reveals that the student has written the identity $\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}$ as $\frac{{{\sin }^{2}}x}{\sin x}\,-\,\frac{{{\cos }^{2}}x}{\cos x}$.
So, this is wrong because the student has split the identity $\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}$ into two separate and irrelevant identities $\frac{{{\sin }^{2}}x}{\sin x}\,\,\text{ and }\,\,\frac{{{\cos }^{2}}x}{\cos x}$.
On solving the equation, he ignored the positive relation between the identities.
Hence, the equation solved by the student is wrong.
