Question : If $A=a \operatorname{cosec} \theta+b \cot \theta$ and $B=a \cot \theta+b \operatorname{cosec} \theta$, then what is the value of $A^2-B^2$?
Option 1: $ab$
Option 2: $(a+b)$
Option 3: $\left(a^2-b^2\right)$
Option 4: $(a-b)$
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Correct Answer: $\left(a^2-b^2\right)$
Solution :
Given, $A=a \operatorname{cosec} \theta+b \cot \theta$ and $B=a \cot \theta+b \operatorname{cosec} \theta$
To find, $A^2-B^2$
Let us find the value of $A^2$ and $B^2$ first,
$A^2 = (a \operatorname{cosec} \theta+b \cot \theta)^2$
$A^2 = a^2 \operatorname{cosec}^2\theta + b^2\cot^2\theta + 2ab\operatorname{cosec}\theta\cot\theta$
Now, $B^2 = (a \cot \theta+b \operatorname{cosec} \theta)^2$
$B^2 = a^2 \cot^2\theta + b^2\operatorname{cosec}^2\theta + 2ab\cot\theta\operatorname{cosec}\theta$
$⇒ A^2-B^2 = (a \operatorname{cosec}^2\theta + b^2\cot^2\theta + 2ab\operatorname{cosec}\theta\cot\theta)-(a^2 \cot^2\theta + b^2\operatorname{cosec}^2\theta + 2ab\cot\theta\operatorname{cosec}\theta)$
$= a^2(\operatorname{cosec}^2\theta - \cot^2\theta) - b^2(\operatorname{cosec}^2\theta - \cot^2\theta)$
$= a^2(1)-b^2(1)$
$=a^2-b^2$
Hence, the correct answer is $(a^2 - b^2)$.
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