Q63 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics
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
- A(I) $\rightarrow$ (a), (II) $\rightarrow$ (e), (III) $\rightarrow$ (d), (IV) $\rightarrow$ (c)
- B(I) $\rightarrow$ (a), (II) $\rightarrow$ (c), (III) $\rightarrow$ (b), (IV) $\rightarrow$ (e)
- C(I) $\rightarrow$ (a), (II) $\rightarrow$ (d), (III) $\rightarrow$ (c), (IV) $\rightarrow$ (b)
- Correct answerD(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 $\rightarrow d$ ]
Hence, the correct answer is option: 'D'.