MCQ
The value of $2\sin^2\text{B}+4\cos(\text{A+B})\sin\text{A}\sin\text{B}+\cos^2(\text{A+B})$ is:
- A$0$
- B$\cos3\text{A}$
- C$\cos2\text{A}$
- DNone of these
Solution:
We have,
$2\sin^2\text{B}+4\cos(\text{A+B})\sin\text{A}\sin\text{B}+\cos2\text{A+B}$
$=1-\cos2\text{B}+\cos^2(\text{A+B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}$
$=1+(\cos2(\text{A+B})-\cos2\text{B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}$
$=1-2\sin\text{A}\sin(\text{A+2B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}\\ \ \ \ \ \ \ \ \ \Big[\because\cos\text{C}\cos\text{D}=-2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}\Big]$
$=1-2\sin\text{A}[\sin(\text{A}+2\text{B})-2\sin\text{B}-2\cos(\text{A+B})]$
$=1-2\sin\text{A}[\sin(\text{A}+2\text{B})-\big\{\sin(\text{B+A+B})+\sin(\text{B}-\text{(A+B)})\big\}\\ \ \ \ \ \ [\because2\sin\text{C}\cos\text{D}=\sin(\text{C+D})+\sin(\text{C}-\text{D})]$
$=1-2\sin\text{A}\big[\sin(\text{A+2B})-\big\{\sin(\text{A+2B})+\sin(-\text{A})\big\}\big]$
$=1-2\sin\text{A}[\sin\text{A}]$
$=1-2\sin^2\text{A}$
$=\cos^2\text{A}$
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