The equation sec^2 theta = frac{4xy}{(x + y)^ | vedclass

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Question

(A) We know that for any real angle $	heta$,$\cos^2 	heta \le 1$,which implies $\sec^2 	heta \ge 1$.Given the equation $\sec^2 	heta = \frac{4xy}{

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$x = y$

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(A) We know that for any real angle $	heta$,$\cos^2 	heta \le 1$,which implies $\sec^2 	heta \ge 1$.Given the equation $\sec^2 	heta = \frac{4xy}{(x + y)^2}$,we must have $\frac{4xy}{(x + y)^2} \ge 1$.Since $(x + y)^2$ is always non-negative,we can multiply both sides by $(x + y)^2$ to get $4xy \ge (x + y)^2$.Expanding the right side,we get $4xy \ge x^2 + 2xy + y^2$.Rearranging the terms,we get $0 \ge x^2 - 2xy + y^2$,which simplifies to $(x - y)^2 \le 0$.Since the square of any real number cannot be negative,the only possibility is $(x - y)^2 = 0$,which means $x = y$.

The equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$ is only possible when

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