The Value of \( \frac{2(\sin2x+2\cos^2x-1)}{\cos x-\sin x-\cos3x+\sin3x} \)
Question
Find the value of
\[ \frac{2(\sin2x+2\cos^2x-1)} {\cos x-\sin x-\cos3x+\sin3x} \]
(a) \(\cos x\)
(b) \(\sec x\)
(c) \(\csc x\)
(d) \(\sin x\)
Solution
First simplify the numerator:
\[ 2\cos^2x-1=\cos2x \]
Therefore,
\[ 2(\sin2x+2\cos^2x-1) = 2(\sin2x+\cos2x) \]
Now simplify the denominator:
\[ \cos x-\cos3x = -2\sin\frac{4x}{2}\sin\frac{-2x}{2} = 2\sin2x\sin x \]
\[ \sin3x-\sin x = 2\cos\frac{4x}{2}\sin\frac{2x}{2} = 2\cos2x\sin x \]
Hence,
\[ \cos x-\sin x-\cos3x+\sin3x \]
\[ =(\cos x-\cos3x)+(\sin3x-\sin x) \]
\[ =2\sin x(\sin2x+\cos2x) \]
Substituting,
\[ \frac{2(\sin2x+\cos2x)} {2\sin x(\sin2x+\cos2x)} \]
\[ =\frac{1}{\sin x} \]
\[ =\csc x \]
Final Answer
\[ \boxed{\csc x} \]
Hence, the correct option is (c) \(\csc x\).