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If $\sin \theta+\cos \theta=p$ and $\tan \theta+\cot \theta=q$, then $q\left(p^2-1\right)$ is equal to
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Verified Answer
The correct answer is:
$2$
We have given,
From Eq. (i)
$\begin{aligned}
& (\sin \theta+\cos \theta)^2=p^2 \\
& \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=p^2 \\
& \quad\left[\because \sin ^2 \theta+\cos ^2 \theta \text { and } \sin 2 \theta=2 \sin \theta \cos \theta\right] \\
& 1+\sin 2 \theta=p^2
\end{aligned}$
From Eq. (ii), $\tan \theta+\cot \theta=q$
$\begin{aligned}
& \tan \theta+\frac{1}{\tan \theta}=q \Rightarrow \frac{\tan ^2 \theta+1}{\tan \theta}=q \\
& \frac{\tan ^2 \theta+1}{2 \tan \theta}=\frac{q}{2} \quad\left[\because \operatorname{cosec} 2 \theta=\frac{\tan ^2 \theta+1}{2 \tan \theta}\right] \\
& \therefore \quad \operatorname{cosec} 2 \theta=\frac{q}{2} \\
& \sin 2 \theta=\frac{2}{q}
\end{aligned}$
$p^2-1=\frac{2}{q}$ [from eq. (iii)]
$q\left(p^2-1\right)=2$
From Eq. (i)
$\begin{aligned}
& (\sin \theta+\cos \theta)^2=p^2 \\
& \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=p^2 \\
& \quad\left[\because \sin ^2 \theta+\cos ^2 \theta \text { and } \sin 2 \theta=2 \sin \theta \cos \theta\right] \\
& 1+\sin 2 \theta=p^2
\end{aligned}$
From Eq. (ii), $\tan \theta+\cot \theta=q$
$\begin{aligned}
& \tan \theta+\frac{1}{\tan \theta}=q \Rightarrow \frac{\tan ^2 \theta+1}{\tan \theta}=q \\
& \frac{\tan ^2 \theta+1}{2 \tan \theta}=\frac{q}{2} \quad\left[\because \operatorname{cosec} 2 \theta=\frac{\tan ^2 \theta+1}{2 \tan \theta}\right] \\
& \therefore \quad \operatorname{cosec} 2 \theta=\frac{q}{2} \\
& \sin 2 \theta=\frac{2}{q}
\end{aligned}$
$p^2-1=\frac{2}{q}$ [from eq. (iii)]
$q\left(p^2-1\right)=2$
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