The equation (a + b)^2 = 4ab sin^2 theta is p | vedclass

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(B) We are given the equation $(a + b)^2 = 4ab \sin^2 	heta$.Since the range of the sine function is $[-1, 1]$,the value of $\sin^2 	heta$ must sati

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$a = b$

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(B) We are given the equation $(a + b)^2 = 4ab \sin^2 	heta$.Since the range of the sine function is $[-1, 1]$,the value of $\sin^2 	heta$ must satisfy $0 \le \sin^2 	heta \le 1$.Therefore,we have $\sin^2 	heta = \frac{(a + b)^2}{4ab} \le 1$.This implies $(a + b)^2 \le 4ab$.Rearranging the terms,we get $(a + b)^2 - 4ab \le 0$.Expanding the square,we get $a^2 + 2ab + b^2 - 4ab \le 0$,which simplifies to $a^2 - 2ab + b^2 \le 0$.This is equivalent to $(a - b)^2 \le 0$.Since the square of any real number cannot be negative,the only possibility is $(a - b)^2 = 0$,which implies $a = b$.

The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

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