CBSE · Class 10 · Mathematics · Introduction to TrigonometryEvaluate the following expression by substituting standard trigonometric values: $\frac{\cos^2 45^\circ}{\sec 30^\circ + \csc 30^\circ}$.
Step-by-Step Evaluation
Given Expression: $$\frac{\cos^2 45^\circ}{\sec 30^\circ + \csc 30^\circ}$$
Step 1: Identify the standard trigonometric values for the given angles.
- $\cos 45^\circ = \frac{1}{\sqrt{2}}$
- $\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}}$
- $\csc 30^\circ = \frac{1}{\sin 30^\circ} = 2$
Step 2: Substitute these values into the given expression.\nNumerator: $\cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$ \nDenominator: $\sec 30^\circ + \csc 30^\circ = \frac{2}{\sqrt{3}} + 2$ \nSo the expression becomes: $$\frac{\frac{1}{2}}{\frac{2}{\sqrt{3}} + 2}$$
Step 3: Simplify the denominator.\nFind a common denominator for the terms in the denominator: $$\frac{2}{\sqrt{3}} + 2 = \frac{2 + 2\sqrt{3}}{\sqrt{3}} = \frac{2(1 + \sqrt{3})}{\sqrt{3}}$$
Step 4: Substitute the simplified denominator back into the main fraction. $$\frac{\frac{1}{2}}{\frac{2(1 + \sqrt{3})}{\sqrt{3}}} = \frac{1}{2} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})} = \frac{\sqrt{3}}{4(1 + \sqrt{3})}$$
Step 5: Rationalize the denominator.\nMultiply the numerator and the denominator by the conjugate of $(1 + \sqrt{3})$, which is $(\sqrt{3} - 1)$ or $(1 - \sqrt{3})$: $$\frac{\sqrt{3}}{4(1 + \sqrt{3})} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3}(\sqrt{3} - 1)}{4(\sqrt{3} + 1)(\sqrt{3} - 1)}$$ \nUsing the algebraic identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
- Denominator: $4((\sqrt{3})^2 - (1)^2) = 4(3 - 1) = 4(2) = 8$
- Numerator: $\sqrt{3}(\sqrt{3} - 1) = 3 - \sqrt{3}$
Step 6: Final Result $$\frac{3 - \sqrt{3}}{8}$$ \nThus, the evaluated value of the expression is $\frac{3 - \sqrt{3}}{8}$.