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Q2(iii):
Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$
Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$
Solution :
Given: The trigonometric equation $\sin 2A = 2 \sin A$.
To find: The value of $A$ for which the given equation holds true, choosing from the standard options usually provided in this context: (A) $0^\circ$, (B) $30^\circ$, (C) $45^\circ$, (D) $60^\circ$.
Visual Representation:
Step 1: Testing Option (A) where $A = 0^\circ$
Substitute $A = 0^\circ$ into the Left Hand Side (LHS) of the equation:
LHS $= \sin 2A = \sin(2 \times 0^\circ) = \sin 0^\circ$
[Since the value of $\sin 0^\circ = 0$ from trigonometric ratio tables]
LHS $= 0$
Now, substitute $A = 0^\circ$ into the Right Hand Side (RHS) of the equation:
RHS $= 2 \sin A = 2 \sin 0^\circ$
[Since $\sin 0^\circ = 0$]
RHS $= 2 \times 0 = 0$
Since LHS = RHS, the equation is true for $A = 0^\circ$.
Step 2: Testing Option (B) where $A = 30^\circ$
LHS $= \sin 2(30^\circ) = \sin 60^\circ$
[Using the standard value $\sin 60^\circ = \frac{\sqrt{3}}{2}$]
LHS $= \frac{\sqrt{3}}{2}$
RHS $= 2 \sin 30^\circ$
[Using the standard value $\sin 30^\circ = \frac{1}{2}$]
RHS $= 2 \times \frac{1}{2} = 1$
Since $\frac{\sqrt{3}}{2} \neq 1$, the equation is false for $A = 30^\circ$.
Step 3: Testing Option (C) where $A = 45^\circ$
LHS $= \sin 2(45^\circ) = \sin 90^\circ$
[Since $\sin 90^\circ = 1$]
LHS $= 1$
RHS $= 2 \sin 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
Since $1 \neq \sqrt{2}$, the equation is false for $A = 45^\circ$.
Step 4: Testing Option (D) where $A = 60^\circ$
LHS $= \sin 2(60^\circ) = \sin 120^\circ$
[Using the identity $\sin(180^\circ - \theta) = \sin \theta$, $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$]
LHS $= \frac{\sqrt{3}}{2}$
RHS $= 2 \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Since $\frac{\sqrt{3}}{2} \neq \sqrt{3}$, the equation is false for $A = 60^\circ$.
Conclusion: Comparing the results, the equation $\sin 2A = 2 \sin A$ holds true only when $A = 0^\circ$.
Final Answer: The correct option is (A) $0^\circ$.
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(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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