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Q1(iv):
Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
Solution :
Given: The trigonometric expression $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$.
To Find: The numerical value of the given expression.
Step 1: Identify the values of the trigonometric ratios.
Using the standard trigonometric table for specific angles, we have:
$\sin 30^\circ = \frac{1}{2}$
$\tan 45^\circ = 1$
$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\cos 60^\circ = \frac{1}{2}$
$\cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1} = 1$
Step 2: Substitute the values into the expression.
Substituting the values identified in Step 1 into the given expression:
$\text{Expression} = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$
Step 3: Simplify the numerator and the denominator.
For the numerator: $\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{3\sqrt{3} - 4}{2\sqrt{3}}$
For the denominator: $\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{4 + 3\sqrt{3}}{2\sqrt{3}}$
Step 4: Perform the division of the fractions.
$\text{Expression} = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{3\sqrt{3} + 4}{2\sqrt{3}}}$
Since the denominators are identical, they cancel out:
$\text{Expression} = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$
Step 5: Rationalize the denominator.
To rationalize, multiply the numerator and denominator by the conjugate of the denominator, which is $(3\sqrt{3} - 4)$:
$\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4} = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - (4)^2}$
Applying the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$ in the numerator and $a^2 - b^2$ in the denominator:
Numerator: $(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2 = (9 \times 3) - 24\sqrt{3} + 16 = 27 - 24\sqrt{3} + 16 = 43 - 24\sqrt{3}$
Denominator: $(3\sqrt{3})^2 - (4)^2 = 27 - 16 = 11$
Thus, the expression simplifies to:
$\frac{43 - 24\sqrt{3}}{11}$
Final Answer: \frac{43 - 24\sqrt{3}}{11}
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(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
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