(A) We know that for any real $\theta$,$\sec^2 \theta \ge 1$.Given the equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$,it must satisfy $\frac{4xy}{(x

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(A) We know that for any real $\theta$,$\sec^2 \theta \ge 1$.Given the equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$,it must satisfy $\frac{4xy}{(x + y)^2} \ge 1$.Since $(x + y)^2 > 0$ (assuming $x, y \neq 0$),we have $4xy \ge (x + y)^2$.$4xy \ge x^2 + 2xy + y^2$.$0 \ge x^2 - 2xy + y^2$.$0 \ge (x - y)^2$.Since the square of a real number cannot be negative,$(x - y)^2$ must be $0$.Therefore,$x - y = 0$,which implies $x = y$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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

IIT 1966 All IIT

A
$x = y$
B
$x < y$
C
$x > y$
D
None of these

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Trigonometrical Ratios, Functions and Identities

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Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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