Question

If $\sin \theta+\cos \theta=p$ and $\tan \theta+\cot \theta=q$, then $q\left(p^{2}-1\right)$ is equal to

• A. $\frac{1}{2}$
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (B)

Given that, $\sin \theta+\cos \theta=p$ .....(i)
$\tan \theta+\cot \theta=q$ ......(ii)
From Eq. (i) $(\sin \theta+\cos \theta)^{2}=p^{2}$
$\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=p^{2}$
$\Rightarrow 1+2 \sin \theta \cos \theta=p^{2} \left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\Rightarrow 1+\sin 2 \theta=p^{2} [\because \sin 2 \theta=2 \sin \theta \cos \theta]$
$\Rightarrow \sin 2 \theta=p^{2}-1$ ......(iii)
From Eq. (ii), $\tan \theta+\cot \theta=q$
$\Rightarrow \tan \theta+\frac{1}{\tan \theta}=q \Rightarrow \frac{\tan ^{2} \theta+1}{\tan \theta}=q$
$\Rightarrow \frac{\tan ^{2} \theta+1}{2 \tan \theta}=\frac{q}{2} \left[\because \operatorname{cosec} 2 \theta=\frac{\tan ^{2} \theta+1}{2 \tan \theta}\right]$
$\therefore \operatorname{cosec} 2 \theta=\frac{q}{2}$
$\Rightarrow \frac{1}{\sin 2 \theta}=\frac{q}{2}$
$\Rightarrow \frac{1}{p^{2}-1}=\frac{q}{2}$ [from Eq.(iii)]
$\Rightarrow $ $2=q\left(p^{2}-1\right)$
$\Rightarrow q\left(p^{2}-1\right)=2$