Q61 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
If $\sin x+\sin y=\alpha, \cos x+\cos y=\beta$ then $\operatorname{cosec}(x+y)=$
- A$\frac{\beta^2-\alpha^2}{\beta^2+\alpha^2}$
- B$\frac{2 \alpha \beta}{\beta^2-\alpha^2}$
- Correct answerC$\frac{\alpha^2+\beta^2}{2 \alpha \beta}$
- D$\frac{2 \alpha \beta}{\beta^2+\alpha^2}$
Solution
$\begin{aligned}
& \text { Since } \sin x+\sin y=\alpha \\
& \Rightarrow 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\alpha
\end{aligned}$
Since, $\cos x+\cos y=\beta$ $\begin{aligned} & \Rightarrow 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\beta \\ & \tan \left(\frac{x+y}{2}\right)=\frac{\alpha}{\beta} \Rightarrow \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\ & =\frac{2 \frac{\alpha}{\beta}}{1+\frac{\alpha^2}{\beta^2}}=\frac{2 \alpha \beta}{\alpha^2+\beta^2} \Rightarrow \operatorname{cosec}(x+y)=\frac{\alpha^2+\beta^2}{2 \alpha \beta} \end{aligned}$
Since, $\cos x+\cos y=\beta$ $\begin{aligned} & \Rightarrow 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\beta \\ & \tan \left(\frac{x+y}{2}\right)=\frac{\alpha}{\beta} \Rightarrow \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\ & =\frac{2 \frac{\alpha}{\beta}}{1+\frac{\alpha^2}{\beta^2}}=\frac{2 \alpha \beta}{\alpha^2+\beta^2} \Rightarrow \operatorname{cosec}(x+y)=\frac{\alpha^2+\beta^2}{2 \alpha \beta} \end{aligned}$
Hence, the correct answer is option: 'C'.