Statement-1 (A): Let a, b be non-zero real numbers. Then, ^2 =4 a… - Vidyadip

MCQ

Statement-1 (A): Let $a, b$ be non-zero real numbers. Then, $\sec ^2 \theta=\frac{4 a b}{(a+b)^2}$ is true if and only if $a=b$.
Statement-2 (R): $\sec ^2 \theta \geq 1$ for $0 \leq \theta<90^{\circ}$.

✓
Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
B
Statement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
C
Statement-1 is True, Statement-2 is False.
D
Statement-1 is False, Statement-2 is True.

✓

Answer

Correct option: A.

Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.

(A)Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
In a right triangle if $\theta$ is one of the acute angles, then
$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }} \geq 1 \Rightarrow \sec ^2 \theta \geq 1 \quad[\because$ Hypotenuse $\geq$ Base $]$
So, statement-2 is true.
Now, $\quad \sec ^2 \theta=\frac{4 a b}{(a+b)^2}$
$\Rightarrow \quad \frac{4 a b}{(a+b)^2} \geq 1 \quad\left[\because \sec ^2 \theta \geq 1\right]$
$\begin{array}{ll}\Rightarrow & 1-\frac{4 a b}{(a+b)^2} \geq 0 \\ \Rightarrow & \frac{(a+b)^2-4 a b}{(a+b)^2} \leq 0\end{array}$
$\Rightarrow \quad \frac{(a-b)^2}{(a+b)^2} \leq 0 \Rightarrow\left(\frac{a-b}{a+b}\right)^2 \leq 0 \Rightarrow\left(\frac{a-b}{a+b}\right)^2=0 \quad\left[\because\left(\frac{a-b}{a+b}\right)^2\right.$ cannot be negative $]$
$
\Rightarrow \quad \frac{a-b}{a+b}=0 \Rightarrow a-b=0 \Rightarrow a=b .
$
So, statement-1 is true and slatement-2 is a correct explanation for statement-1.

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