Answer
The equation of the tangent line to the curve at the given point $(\pi/2, \pi/4)$ is,
$y = \frac{1}{2}x$
Work Step by Step
Given that,
$ysin2x = xcos2y$, $(\pi/2, \pi/4)$
After differentiating both side
we get,
⇒ $y.cos2x.2+sin2x.y' = x.(-sin2y).2y'+cos2y.1$
⇒ $2ycos2x+y'sin2x = -2xy'sin2y+cos2y$
⇒ $y'sin2x+2xy'sin2y= cos2y-2ycos2x$
⇒ $y'(sin2x+2xsin2y)= cos2y-2ycos2x$
⇒ $y'= \frac{cos2y-2ycos2x}{sin2x+2xsin2y}$
At the given point $(\pi/2, \pi/4)$ ,
$y' = \frac{cos(2.\frac{\pi}{4})-2.\frac{\pi}{4}cos(2.\frac{\pi}{2})}{sin(2.\frac{\pi}{2})+2.\frac{\pi}{2}sin(2.\frac{\pi}{4})}$
⇒ $y' = \frac{cos\frac{\pi}{2}-\frac{\pi}{2}cos\pi}{sin\pi+ \pi sin\frac{\pi}{2}}$
⇒ $y' = \frac{0 -\frac{\pi}{2}.(-1)}{0 + \pi.1}$
⇒ $y' = \frac{\frac{\pi}{2}}{ \pi}$
⇒ $y' = \frac{1}{2}$
So, an equation of the tangent line to the curve at the given point $(\pi/2, \pi/4)$ is,
$y-\frac{\pi}{4} = \frac{1}{2}(x-\frac{\pi}{2})$
⇒ $y = \frac{1}{2}x - \frac{\pi}{4} + \frac{\pi}{4}$
⇒ $y = \frac{1}{2}x$
