# If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2 =

Question:

If $a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta-a \operatorname{cosec} \theta=q$, then $p^{2}-q^{2}=$

(a) $a^{2}-b^{2}$

(b) $b^{2}-a^{2}$

(C) $a^{2}+b^{2}$

(d) $b-a$

Solution:

Given:

$a \cot \theta+b \operatorname{cosec} \theta=p$

$b \cot \theta+a \operatorname{cosec} \theta=q$

Squaring both the equations and then subtracting the second from the first, we have

$(p)^{2}-(q)^{2}=(a \cot \theta+b \operatorname{cosec} \theta)^{2}-(b \cot \theta+a \operatorname{cosec} \theta)^{2}$

$=\left(a^{2} \cot ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta+2 \cdot a \cot \theta \cdot b \operatorname{cosec} \theta\right)-\left(b^{2} \cot ^{2} \theta+a^{2} \operatorname{cosec}^{2} \theta+2 b \cot \theta \cdot a \operatorname{cosec} \theta\right)$

$=a^{2} \cot ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta+2 a b \cot \theta \operatorname{cosec} \theta-b^{2} \cot ^{2} \theta-a^{2} \operatorname{cosec}^{2} \theta-2 a b \cot \theta \operatorname{cosec} \theta$

$=a^{2} \cot ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta-b^{2} \cot ^{2} \theta-a^{2} \operatorname{cosec}^{2} \theta$

$\Rightarrow\left(b^{2} \operatorname{cosec}^{2} \theta-b^{2} \cot ^{2} \theta\right)+\left(-a^{2} \operatorname{cosec}^{2} \theta+a^{2} \cot ^{2} \theta\right)=p^{2}-q^{2}$

$\Rightarrow b^{2}\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)-a^{2}\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)=p^{2}-q^{2}$

$\Rightarrow b^{2}(1)-a^{2}(1)=p^{2}-q^{2}$

$\Rightarrow b^{2}-a^{2}=p^{2}-q^{2}$

$\Rightarrow p^{2}-q^{2}=b^{2}-a^{2}$

Hence, the correct option is (b).