CBSE Class 10 Mathematics Basic 2025: question 24

Hints

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Step-by-step solution

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1. Using $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, $\sec 30^{\circ }=\frac{2}{\sqrt{3}}$, $\mathrm{cosec}30^{\circ }=2$: $\frac{\cos 45^{\circ }}{\sec 30^{\circ }+\mathrm{cosec}30^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$.

   1½ marks

2. $=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2+2\sqrt{3}}=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$; rationalising, $=\frac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}=\frac{3\sqrt{2}-\sqrt{6}}{8}$.

   ½ mark

Answer: $\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$ or $\frac{3\sqrt{2}-\sqrt{6}}{8}$

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Relationships Between Trigonometric Ratios