CBSE Class 10 Mathematics Standard 2024: question 25

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Use the standard values: $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, $\sec 30^{\circ }=\frac{2}{\sqrt{3}}$, $\mathrm{cosec}30^{\circ }=2$. So the expression is $\frac{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}}{\frac{2}{\sqrt{3}}+2}$.

   1½ marks

2. Numerator $=\frac{2+\sqrt{6}}{2\sqrt{2}}$ and denominator $=\frac{2+2\sqrt{3}}{\sqrt{3}}$. Their quotient is $\frac{\sqrt{3}(2+\sqrt{6})}{2\sqrt{2}\cdot 2(1+\sqrt{3})}=\frac{2\sqrt{3}+3\sqrt{2}}{4\sqrt{2}(1+\sqrt{3})}$. Rationalising gives $\frac{3\sqrt{3}+3\sqrt{2}-\sqrt{6}-3}{8}$.

   ½ mark

Answer: $\frac{2\sqrt{3}+3\sqrt{2}}{4\sqrt{2}(1+\sqrt{3})}$ (= $\frac{3\sqrt{3}+3\sqrt{2}-\sqrt{6}-3}{8}$)

Revise the chapter

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