Question

If $(\sin \theta ,\cos \theta )$ and $(3,2)$ lies on the same side of the line $x+y=1$, then $\theta$ lies between

• A. $\left(0,\frac{\pi }{2}\right)$
• B. $(0,\pi )$
• C. $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$
• D. $\left(0,\frac{\pi }{4}\right)$

Answer 1

Answer: (A)

Given points are $(\sin \theta ,\cos \theta )$ and $(3,2)$ and a line
$x+y-1=\mathrm{0.....}$(i)
Since, $(3,2)$ lies on Eq. (i)
$\therefore 3+2-1>0$
and $(\sin \theta ,\cos \theta )$ lies in Eq. (i)
$\therefore \sin \theta +\cos \theta -1>0$
$\Rightarrow \sin \theta +\cos \theta >1$
$\Rightarrow \sqrt{2}\left[\sin \left(\theta +\frac{\pi }{4}\right)\right]>1$
$\Rightarrow \sin \left(\theta +\frac{\pi }{4}\right)>\frac{1}{\sqrt{2}}=\sin \left(\frac{\pi }{4}\right)$

$\Rightarrow \frac{\pi }{4}<\theta +\frac{\pi }{4}<\frac{3\pi }{4}\Rightarrow 0<\theta <\frac{\pi }{2}$
Hence, $\theta \in \left(0,\frac{\pi }{2}\right)$