Answer:
Step-by-step explanation:
To simplify the given expression \(\frac{{(\sin(x) + \cos(x))^2}}{{\sin(2x)}} = \csc(2x) + 1\), let's work on both sides of the equation separately.
Starting with the left side:
\[ \frac{{(\sin(x) + \cos(x))^2}}{{\sin(2x)}} \]
Expand the numerator:
\[ \frac{{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)}}{{\sin(2x)}} \]
Now, use the double-angle identity for sine: \(\sin(2x) = 2\sin(x)\cos(x)\):
\[ \frac{{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)}}{{2\sin(x)\cos(x)}} \]
Combine the terms in the numerator:
\[ \frac{{\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x)}}{{2\sin(x)\cos(x)}} \]
Now, use the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\):
\[ \frac{{1 + 2\sin(x)\cos(x)}}{{2\sin(x)\cos(x)}} \]
Factor out 2 in the numerator:
\[ \frac{{\frac{1}{2} + \sin(x)\cos(x)}}{{\sin(x)\cos(x)}} \]
Cancel out \(\sin(x)\cos(x)\) in the numerator and denominator:
\[ \frac{{\frac{1}{2}}}{{1}} = \frac{1}{2} \]
So, the left side simplifies to \(\frac{1}{2}\).
Now, let's examine the right side:
\[ \csc(2x) + 1 \]
Recall that \(\csc(2x) = \frac{1}{\sin(2x)}\), so:
\[ \frac{1}{{\sin(2x)}} + 1 \]
Now, substitute the double-angle identity \(\sin(2x) = 2\sin(x)\cos(x)\):
\[ \frac{1}{{2\sin(x)\cos(x)}} + 1 \]
Combine the terms over a common denominator:
\[ \frac{1 + 2\sin(x)\cos(x)}{{2\sin(x)\cos(x)}} \]
Now, cancel out \(\sin(x)\cos(x)\) in the numerator and denominator:
\[ \frac{1}{2} \]
So, the right side also simplifies to \(\frac{1}{2}\).
Therefore, the given expression \(\frac{{(\sin(x) + \cos(x))^2}}{{\sin(2x)}}\) is equivalent to \(\csc(2x) + 1\).
