Question

If $x=\sin \theta \left|\sin \theta \right|,y=\cos \theta \left|\cos \theta \right|,$ $\frac{99\pi }{2}\le \theta \le \frac{100\pi }{2},$ then

• A. $x-y=1$
• B. $x+y=-1$
• C. $x+y=1$
• D. $y-x=1$

Answer 1

The correct option is D $y-x=1$

Solution -

In $\theta ϵ[\frac{99\pi }{2},\frac{100\pi }{2}]$

$\frac{99\pi }{2}=48\pi +\frac{3\pi }{2},$

$\frac{100\pi }{2}=48\pi +2\pi$

$\theta ϵ[\frac{3\pi }{2},2\pi ]$ $sin\theta =-ive$

$cos\theta =+ive$

$x=-si{n}^2\theta$

$y=co{s}^2\theta$

$y-x=co{s}^2\theta +si{n}^2\theta =1$

$y-x=1$

D is correct