Trigonometrical Equations 3 Question 9
10. $\sec ^{2} \theta=\frac{4 x y}{(x+y)^{2}}$ is true if and only if
(1996, 1M)
(a) $x=y \neq 0$
(b) $x=y, x \neq 0$
(c) $x=y$
(d) $x \neq 0, y \neq 0$
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Answer:
Correct Answer: 10. (a, b)
Solution:
- We know that, $\sec ^{2} \theta \geq 1$
$ \begin{array}{lc} \Rightarrow & \frac{4 x y}{(x+y)^{2}} \geq 1 \\ \Rightarrow & 4 x y \geq(x+y)^{2} \\ \Rightarrow & (x+y)^{2}-4 x y \leq 0 \\ \Rightarrow & (x-y)^{2} \leq 0 \\ \Rightarrow & x-y=0 \\ \Rightarrow & x=y \end{array} $
Therefore, $x+y=2 x$
[add $x$ both sides]
But $x+y \neq 0$ since it lies in the denominator,
$\Rightarrow \quad 2 x \neq 0$
$\Rightarrow \quad x \neq 0$
Hence, $x=y, x \neq 0$ is the answer.
Therefore, (a) and (b) are the answers.
