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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles4 Marks
Question
$\text{If} \ \cos\text{A}+\sin\text{B}=\text{m}$ and $\sin\text{A}+\cos\text{B}=\text{n},$ prove that $2\sin\text{(A}+\text{B)}=\text{m}^2+\text{n}^2-2. $
✓

Answer

We have,
$\cos\text{A}+\sin\text{B}=\text{m}\ \text{and}\ \sin\text{A}+\cos\text{B}=\text{n}$
$=\text{m}^2+\text{n}^2-2$
$=(\cos\text{A}+\sin\text{B})^2+(\sin\text{A}+\cos\text{B)}^2-2$
$ =\cos^2\text{A}+\sin^2\text{B}+2\cos\text{A}\sin\text{B}+\sin^2\text{A}\cos^2\text{B}+2\sin\text{A}\cos\text{B}-2$
$ =(\sin^2\text{A}+\cos^2\text{A})+(\sin^2\text{B}+\cos^2\text{B})\\\ +2\cos\text{A}\sin\text{B}+\sin^2\text{A}\cos \text{B}+2\sin\text{A}\cos\text{B}-2 $
$=1+1+2\cos\text{A}\sin\text{B}+2\sin\text{A}\cos\text{B}-2$
$=2+2(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B)}-2$
$=2(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B)}$
$ =2\sin\text{(A}+\text{B)}$$\big[\because\sin\text{(A}+\text{B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\big]$
$\therefore2\sin\text{(A}+\text{B)}=\text{m}^2+\text{n}^2-2 $
Hence proved.

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