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Q1(v):
Evaluate the following : (v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
Solution :
Given: The trigonometric expression $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.
To Find: The numerical value of the given expression.
Step 1: Identify the values of the trigonometric ratios.
Based on the standard trigonometric table for specific angles, we have:
$\cos 60^\circ = \frac{1}{2}$
$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\tan 45^\circ = 1$
$\sin 30^\circ = \frac{1}{2}$
$\cos 30^\circ = \frac{\sqrt{3}}{2}$
Step 2: Substitute the values into the numerator.
The numerator is $5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ$.
Substituting the values:
$= 5 \left( \frac{1}{2} \right)^2 + 4 \left( \frac{2}{\sqrt{3}} \right)^2 - (1)^2$
$= 5 \left( \frac{1}{4} \right) + 4 \left( \frac{4}{3} \right) - 1$
$= \frac{5}{4} + \frac{16}{3} - 1$
Step 3: Simplify the numerator.
To add the fractions, find the least common multiple (LCM) of 4 and 3, which is 12.
$= \frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{1 \times 12}$
$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12}$
$= \frac{15 + 64 - 12}{12}$
$= \frac{67}{12}$
Step 4: Simplify the denominator.
The denominator is $\sin^2 30^\circ + \cos^2 30^\circ$.
[Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ for any angle $\theta$]:
$\sin^2 30^\circ + \cos^2 30^\circ = 1$
Alternatively, by calculation:
$= \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2$
$= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
Step 5: Calculate the final value.
The expression is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{67/12}{1}$.
$= \frac{67}{12}$
Final Answer: \frac{67}{12}
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(iii): Evaluate the following : (iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$
- Q1(iv): Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
- 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
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