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