Q174 / 270 · NDA 2020 · Mathematics
Paragraph:
Read the following information and answer:
Let $\alpha=\beta=15^{\circ}$.
Question:
What is $\sin \left(\alpha+1^{\circ}\right)+\cos \left(\beta+1^{\circ}\right)$ equal to?
Question: What is $\sin \left(\alpha+1^{\circ}\right)+\cos \left(\beta+1^{\circ}\right)$ equal to?
- A$\sqrt{3} \cos 1^{\circ}+\sin 1^{\circ}$
- B$\sqrt{3} \cos 1^{\circ}-\frac{1}{2} \sin 1^{\circ}$
- Correct answerC$\frac{1}{\sqrt{2}}\left(\sqrt{3} \cos 1^{\circ}+\sin 1^{\circ}\right)$
- D$\frac{1}{2}\left(\sqrt{3} \cos 1^{\circ}+\sin 1^{\circ}\right)$
Solution
$\sin \left(\alpha+1^{\circ}\right)+\cos \left(\beta+1^{\circ}\right)$
$=\sqrt{2}\left\{\frac{1}{\sqrt{2}} \cdot \sin \left(\alpha+1^{\circ}\right)+\frac{1}{\sqrt{2}} \cdot \cos \left(\beta+1^{\circ}\right)\right\}$
$\begin{aligned}=\sqrt{2}\left\{\cos 45^{\circ} \cdot \sin (\alpha+1)+\sin 45^{\circ} \cdot \cos (\alpha+1)\right\} & \\ & (\because \alpha=\beta)\end{aligned}$
$\begin{aligned}
& =\sqrt{2}\left\{\sin \left(\alpha+1+45^{\circ}\right)\right\}=\sqrt{2} \cdot \sin \left(15^{\circ}+1^{\circ}+45^{\circ}\right) \\
& =\sqrt{2} \cdot \sin 61^{\circ} (i)
\end{aligned}$
Again
$\begin{aligned}
& \frac{1}{\sqrt{2}}\left(\sqrt{3} \cdot \cos 1^{\circ}+\sin 1^{\circ}\right) \\
& =\frac{2}{\sqrt{2}}\left(\frac{\sqrt{3}}{2} \cdot \cos 1^{\circ}+\frac{1}{2} \sin 1^{\circ}\right) \\
& =\sqrt{2} \cdot\left(\sin 60^{\circ} \cdot \cos 1^{\circ}+\cos 60^{\circ} \cdot \sin 1^{\circ}\right) \\
& =\sqrt{2}\left(\sin 61^{\circ}\right) (ii) \\
& \text { Thus }(\text { i })=(\text { ii })
\end{aligned}$
$\frac{1}{\sqrt{2}}\left(\sqrt{3} \cdot \cos 1^{\circ}+\sin 1^{\circ}\right)$.
Hence, the correct answer is option: 'C'.