Mathematics — Standard (041) — CBSE Class 10 board question

Given: $\sin\alpha = \dfrac{1}{\sqrt{2}}$ and $\cot\beta = \sqrt{3}$

From the standard table:
$$\sin 45° = \frac{1}{\sqrt{2}} \Rightarrow \alpha = 45°$$
$$\cot 30° = \sqrt{3} \Rightarrow \beta = 30°$$

Now:
$$\csc\alpha = \frac{1}{\sin 45°} = \sqrt{2}$$
$$\csc\beta = \frac{1}{\sin 30°} = 2$$

$$\therefore \csc\alpha + \csc\beta = \sqrt{2} + 2$$

Source: Chapter 8, Section 8.3 (Table 8.1)

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• Identify the angles from Table 8.1: $\sin 45° = \tfrac{1}{\sqrt{2}}$ and $\cot 30° = \sqrt{3}$.
• Use the reciprocal relation $\csc A = \tfrac{1}{\sin A}$ to find each value.
• Examiners award 1 mark for correctly identifying both angles/values and 1 mark for the final answer $\sqrt{2} + 2$.