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

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Given: a cot$\theta$+b cosec$\theta$=P, b cot$\theta$+a 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.cosec$\theta$)^2-(b cot $\theta$+a cosec$\theta$)^2 =(a^2cot^$\theta$+b^2 cosec^2$\theta$+2.a cot$\theta$.b cosec$\theta$)-(b^2 cot^2$\theta$+a^2 cosec^2$\theta$+2 cot$\theta$.a cosec$\theta$) =a^2 cot^2$\theta$+b^2 cosec^2$\theta$+2 ab cot$\theta$ cosec$\theta$-b^2 cot^2$\theta$-a^2cosec^2$\theta$-2ab cot$\theta$cosec$\theta$ $\Rightarrow$a^2 cot^2$\theta$+b^2 cosec^2$\theta$-b^2 cot^2$\theta$-a^2 cosec^2$\theta$ $\Rightarrow$(b^2 cosec^$\theta$-b^2 cot^2 $\theta$)+(-a^2 cosec^2$\theta$+a^2 cot^2$\theta$)=p^2-q^2 $\Rightarrow$b^2(cosec^2$\theta$-cot^2$\theta$)-a^2(cosec^$\theta$-cot^2$\theta$)=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`

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