If $\cos \alpha+\cos \beta=a$ and $\sin \alpha+\sin \beta=b$, then match the items given in List $-A$ with…
If $\cos \alpha+\cos \beta=a$ and $\sin \alpha+\sin \beta=b$, then match the items given in List $-A$ with those of their values in List-B
- (I) $\rightarrow$ (a), (II) $\rightarrow$ (e), (III) $\rightarrow$ (d), (IV) $\rightarrow$ (c)
- (I) $\rightarrow$ (a), (II) $\rightarrow$ (c), (III) $\rightarrow$ (b), (IV) $\rightarrow$ (e)
- (I) $\rightarrow$ (a), (II) $\rightarrow$ (d), (III) $\rightarrow$ (c), (IV) $\rightarrow$ (b)
- (I) $\rightarrow$ (a), (II) $\rightarrow$ (d), (III) $\rightarrow$ (b), (IV) $\rightarrow$ (c)
Solution
(d) $\cos \alpha+\cos \beta=a \Rightarrow a=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$ and $\sin \alpha+\sin \beta=b \Rightarrow b=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$ $\frac{b}{a}=\tan \left(\frac{\alpha+\beta}{2}\right)$
[Hence $(\mathrm{I}) \rightarrow(a)$] Now $\tan (\alpha+\beta)=\frac{2 \tan \left(\frac{\alpha+\beta}{2}\right)}{1-\tan ^2\left(\frac{\alpha+\beta}{2}\right)}=\frac{2 a b}{a^2-b^2}$
[Hence (IV) $\rightarrow(c)$ ]
Since $\tan (\alpha+\beta)=\frac{2 a b}{a^2-b^2}$
$
\sin (\alpha+\beta)=\frac{2 a b}{a^2+b^2}
$
and
[Hence III $\rightarrow b$ ]
$
\cos (\alpha+\beta)=\frac{a^2-b^2}{a^2+b^2}
$
[Hence II $\rightarrow d$ ]
Asked in: AP EAMCET 2023 (15 May Shift 1)
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