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Question
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
Options
0
cos 3A
cos 2A
none of these
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Solution
cos 2A
\[\text{ We have, } \]
\[2 \sin^2 B + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B + \cos2\left( A + B \right)\]
\[ = 1 - \cos2B + \cos2\left( A + B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]
\[ = 1 + \left( \cos2\left( A + B \right) - \cos2B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]
\[ = 1 - 2\text{ sin } A\sin\left( A + 2B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]
\[ \left[ \because \text{ cos } C - \text{ cos } D = - 2\sin\frac{C + D}{2}\sin\frac{C - D}{2} \right]\]
\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - 2\text{ sin } B\cos\left( A + B \right) \right]\]
\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - \left\{ \sin\left( B + A + B \right) + \sin\left( B - \left( A + B \right) \right) \right\} \right] \]
\[ \left[ \because 2\text{ sin } C\text{ cos } D = \sin\left( C + D \right) + \sin\left( C - D \right) \right]\]
\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - \left\{ \sin\left( A + 2B \right) + \sin\left( - A \right) \right\} \right]\]
\[ = 1 - 2\text{ sin } A\left[ \text{ sin } A \right]\]
\[ = 1 - 2 \sin^2 A\]
\[ = \cos2A\]
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