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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
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that $\cos(\alpha+\beta)=\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}$
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Answer

$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$ $=\sin^2\alpha+\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\sin\alpha\sin\beta+2\cos\alpha\cos\beta$ $=2+2\cos(\alpha+\beta)$ $\Rightarrow\text{b}^2+\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2$ $\Rightarrow\text{b}^2+\text{a}^2=\cos^2\alpha+\cos^2\beta-\sin^2\alpha+\sin^2\beta+2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$ $\Rightarrow\text{b}^2+\text{a}^2=\Big(\cos^2\alpha+\sin^2\beta\Big)+\Big(\cos^2\beta-\sin^2\alpha\Big)-2\cos(\alpha+\beta)$ $\Rightarrow\text{b}^2+\text{a}^2=2\cos(\alpha+\beta)+\cos(\alpha-\beta)+2\cos(\alpha-\beta)$ $\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(2+2\cos(\alpha-\beta) )$ $\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2 )$ $\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}=\cos(\alpha+\beta)$

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