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Trigonometric Ratios of Multiple and Submultiple Angles — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios of Multiple and Submultiple 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. $
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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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