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Q1(iii):
Evaluate the following : (iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$
Solution :
Given: The trigonometric expression $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$.
To Find: The numerical value of the given expression.
Visual Representation (Trigonometric Ratios):
Step 1: Identify the values of the trigonometric ratios.
Using the standard trigonometric table for specific angles:
- $\cos 45^\circ = \frac{1}{\sqrt{2}}$
- $\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
- $\text{cosec } 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2$
Step 2: Substitute the values into the expression.
The expression is $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$. Substituting the values:
$\text{Expression} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
Step 3: Simplify the denominator.
To add the terms in the denominator, find a common denominator:
$\frac{2}{\sqrt{3}} + 2 = \frac{2 + 2\sqrt{3}}{\sqrt{3}} = \frac{2(1 + \sqrt{3})}{\sqrt{3}}$
Step 4: Perform the division of fractions.
$\text{Expression} = \frac{1}{\sqrt{2}} \div \frac{2(1 + \sqrt{3})}{\sqrt{3}}$
$\text{Expression} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})}$
$\text{Expression} = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} = \frac{\sqrt{3}}{2(\sqrt{2} + \sqrt{6})}$
Step 5: Rationalize the denominator.
Multiply the numerator and denominator by the conjugate $(\sqrt{6} - \sqrt{2})$:
$\text{Expression} = \frac{\sqrt{3}}{2(\sqrt{6} + \sqrt{2})} \times \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}$
$= \frac{\sqrt{18} - \sqrt{6}}{2((\sqrt{6})^2 - (\sqrt{2})^2)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{2(6 - 2)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{2(4)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{8}$
Final Answer: $\frac{3\sqrt{2} - \sqrt{6}}{8}$
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.2
- Q1(i): Evaluate the following : (i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
- Q1(ii): Evaluate the following : (ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
- Q1(iv): Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
- Q1(v): Evaluate the following : (v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
- Q2(i): Choose the correct option and justify your choice : (i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
- Q2(ii): Choose the correct option and justify your choice : (ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
- Q2(iii): Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$
- Q2(iv): Choose the correct option and justify your choice : (iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$
- Q3: If $\tan (A + B) = \sqrt{3}$ and $\tan (A – B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$.
- Q4(i): State whether the following are true or false. Justify your answer. (i) $\sin (A + B) = \sin A + \sin B$.
- Q4(ii): State whether the following are true or false. Justify your answer. (ii) The value of $\sin \theta$ increases as $\theta$ increases.
- Q4(iii): State whether the following are true or false. Justify your answer. (iii) The value of $\cos \theta$ increases as $\theta$ increases.
- Q4(iv): State whether the following are true or false. Justify your answer. (iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
- Q4(v): State whether the following are true or false. Justify your answer. (v) $\cot A$ is not defined for $A = 0^\circ$.
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
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