Question : If $a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta+a \operatorname{cosec} \theta=q$ then $p^2-q^2$ is equal to.
Option 1: $a^2+b^2$
Option 2: $a^2-b^2$
Option 3: $b^2-a^2$
Option 4: $b-a$
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Correct Answer: $b^2-a^2$
Solution :
Given: $a \cot \theta+b \operatorname{cosec} \theta=p$
Squaring both sides,
We get $(a \cot \theta+b \operatorname{cosec} \theta)^{2} =p^{2}$
⇒ $a^{2} \cot^{2}\theta+b^{2} \operatorname{cosec}^{2}\theta +2ab \cot \theta\operatorname{cosec} \theta =p^{2}$……(i)
And $b \cot \theta+a \operatorname{cosec} \theta=q$
Squaring both sides,
We get $(b \cot \theta+a \operatorname{cosec} \theta)^{2}=q^2$
⇒ $b^2 \cot^{2}\theta+a^2 \operatorname{cosec}^{2}\theta+2ab \cot \theta \operatorname{cosec} \theta=q^2$……(ii)
Now, $p^2- q^2$ = $a^{2} \cot^{2}\theta+b^{2} \operatorname{cosec}^{2}\theta +2ab \cot \theta\operatorname{cosec} \theta- b^2 \cot^{2}\theta-a^2 \operatorname{cosec}^{2}\theta-2ab \cot \theta \operatorname{cosec} \theta = p^2 -q^2$
$\because$ $\operatorname{cosec}^{2}\theta - \cot^{2}\theta = 1$
⇒ $p^2 - q^2 = a^2 (\cot^{2}\theta-\operatorname{cosec}^{2}\theta) + b^2 (\operatorname{cosec}^{2}\theta -\cot^{2}\theta)= a^2 ( – 1) + b^2 (1)$ = $b^2 -a^2$
$\therefore p^2 - q^2 = b^2 -a^2$.
Hence the correct answer is $b^2 -a^2$.
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