SATHEE: Trigonometrical Equations 3 Question 9

Trigonometrical Equations 3 Question 9

10. $\sec^{2} θ = \frac{4 x y}{(x + y)^{2}}$ is true if and only if

(1996, 1M)

(a) $x = y \neq 0$

(b) $x = y, x \neq 0$

(d) $x \neq 0, y \neq 0$

Show Answer

Answer:

Correct Answer: 10. (a, b)

Solution:

We know that, $\sec^{2} θ \geq 1$

$\begin{array}{lc} \Rightarrow & \frac{4 x y}{(x + y)^{2}} \geq 1 \\ \Rightarrow & 4 x y \geq (x + y)^{2} \\ \Rightarrow & (x + y)^{2} - 4 x y \leq 0 \\ \Rightarrow & (x - y)^{2} \leq 0 \\ \Rightarrow & x - y = 0 \\ \Rightarrow & x = y \end{array}$

Therefore, $x + y = 2 x$

[add $x$ both sides]

But $x + y \neq 0$ since it lies in the denominator,

$\Rightarrow 2 x \neq 0$

$\Rightarrow x \neq 0$

Hence, $x = y, x \neq 0$ is the answer.

Therefore, (a) and (b) are the answers.

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Trigonometrical Equations 3 Question 9

10. sec2⁡θ=4xy(x+y)2 is true if and only if

Answer:

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10. $\sec^{2} θ = \frac{4 x y}{(x + y)^{2}}$ is true if and only if