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Q2(iv):
Choose the correct option and justify your choice : (iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$
Choose the correct option and justify your choice : (iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$
Solution :
Given: The trigonometric expression $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$.
To Find: The value of the expression and identify the correct option among the standard trigonometric values.
Step 1: Recall the standard trigonometric ratio for $30^\circ$.
From the trigonometric table for standard angles, we know that:
$\tan 30^\circ = \frac{1}{\sqrt{3}}$
Step 2: Substitute the value into the given expression.
Let the expression be $E$.
$E = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 - \left( \frac{1}{\sqrt{3}} \right)^2}$
Step 3: Simplify the numerator and the denominator.
Numerator: $2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$
Denominator: $1 - \left( \frac{1}{\sqrt{3}} \right)^2 = 1 - \frac{1}{3}$
[Since $(\sqrt{3})^2 = 3$]
Denominator: $\frac{3 - 1}{3} = \frac{2}{3}$
Step 4: Perform the division of the fractions.
$E = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
$E = \frac{2}{\sqrt{3}} \times \frac{3}{2}$
[Multiplying by the reciprocal of the denominator]
$E = \frac{2 \times 3}{2 \times \sqrt{3}}$
$E = \frac{3}{\sqrt{3}}$
Step 5: Rationalize the denominator.
$E = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$E = \frac{3\sqrt{3}}{3}$
$E = \sqrt{3}$
Step 6: Compare the result with standard trigonometric values.
We know that $\tan 60^\circ = \sqrt{3}$.
Therefore, the value of the expression is equivalent to $\tan 60^\circ$.
Final Answer: The value of the expression is $\sqrt{3}$, which corresponds to $\tan 60^\circ$.
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) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
- 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 $A =$
- 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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