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Statement-1 : If e^(itheta)=cos theta +...

Statement-1 : If ` e^(itheta)=cos theta +isintheta` and the value of ` e^(iA).e^(iB).e^(iC)` is equal to -1, where A,B,C are the angles of a triangle.
and
Statement -2 : In any ` triangleABC , A + B +C = 180^(@)`

A

Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -1

B

Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -1

C

Statement -1 is True, Statement -2 is False

D

Statement -1 is Flase, Statement -2 is True

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the two statements provided and establish their validity step by step. ### Step 1: Understanding the first statement The first statement is: \[ e^{iA} \cdot e^{iB} \cdot e^{iC} = -1 \] where \( A, B, C \) are the angles of a triangle. Using the property of exponents, we can combine the left side: \[ e^{iA} \cdot e^{iB} \cdot e^{iC} = e^{i(A + B + C)} \] ### Step 2: Relating angles of a triangle From geometry, we know that the sum of the angles in a triangle is: \[ A + B + C = 180^\circ \] ### Step 3: Substituting the sum of angles Substituting this into our expression gives: \[ e^{i(A + B + C)} = e^{i \cdot 180^\circ} \] ### Step 4: Evaluating \( e^{i \cdot 180^\circ} \) Using Euler's formula: \[ e^{i \theta} = \cos \theta + i \sin \theta \] we can evaluate: \[ e^{i \cdot 180^\circ} = \cos(180^\circ) + i \sin(180^\circ) \] ### Step 5: Finding the values of cosine and sine We know: \[ \cos(180^\circ) = -1 \] \[ \sin(180^\circ) = 0 \] Thus: \[ e^{i \cdot 180^\circ} = -1 + 0i = -1 \] ### Step 6: Conclusion for Statement 1 Since we have shown that: \[ e^{iA} \cdot e^{iB} \cdot e^{iC} = -1 \] is true when \( A + B + C = 180^\circ \), the first statement is correct. ### Step 7: Understanding the second statement The second statement is: "In any triangle, \( A + B + C = 180^\circ \)." This is a fundamental property of triangles and is always true. ### Final Conclusion Both statements are correct, and Statement 2 provides a valid explanation for Statement 1.
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