If x cos(a+y) = cos y, then prove that dydx=cos2(a+y)sina
Hence show that sin a d2ydx2+sin2(a+y)dydx=0
OR Find dydx if y=sin−1[6x−4√1−4x25].
Given cos y = x cos(a + y) ⇒x=cos ycos(a+y)
Diff. w.r.t. y both sides, we get dxdy=cos(a+y)(−sin y)−cos y(−sin(a+y))cos2(a+y)
⇒dxdy=cos y sin(a+y)−cos(a+y)sin ycos2(a+y)⇒dxdy=sin(a+y−y)cos2(a+y) ∴dydx=cos2(a+y)sin a.
Now sin adydx=cos2(a+y) ⇒sin ad2ydx2=−2cos(a+y)sin(a+y)dydx
⇒sin ad2ydx2=−sin2a+ydydx ∴sin ad2ydx2+sin2(a+y)dydx=0.
OR
We have y=sin−1[6x−4√1−4x25]. Put 2x=sin θ⇒θ=sin−12x.....(i)
∴y=sin−1[3 sin θ−4√1−sin2 θ5]. ⇒y=sin−1[35sin θ−45cos θ]Let 35=cos α⇒sin α=45...(ii) y=sin−1[cos α sin θ−sin α cos θ]⇒y=sin−1[sin(θ−α)]=θ−α ⇒y=sin−12x−sin−145[By(i)and(ii)]∴dydx=2√1−4x2−0=2√1−4x2