Question

Let $(a-b\cos y)(a+b\cos x)=a^2-b^2$ and $\frac{dy}{dx}=\frac{\sin xf\left(y\right))}{{(a+b\cos x)}^2}$. If $a^2-b^2=192,thenf\left(\pi /2\right).$

Answer 1

Expanding the above expression, we get

$a^2+ab\left(\cos \right(x)-\cos (y\left)\right)-b^2\cos x\cos y=a^2-b^2$

$a\left[\cos \right(x)-\cos (y\left)\right]=b\left[\cos \right(x\left)\cos \right(y)-1]$

Differentiating with respect to x.

$a\left[\sin \right(y).y^{\prime }-\sin (x\left)\right]=-b\left[\sin \right(x\left)\cos \right(y)+\cos (x\left)\sin \right(y).y^{\prime }]$

$y^{\prime }\left[a\sin \right(y)+b\cos (x\left)\sin \right(y\left)\right]=a\sin \left(x\right)-b\sin \left(x\right)\cos \left(y\right)$

$y^{\prime }=\frac{a\sin \left(x\right)-b\sin \left(x\right)\cos \left(y\right)}{a\sin \left(y\right)+b\cos \left(x\right)\sin \left(y\right)}$

$=\frac{\sin \left(x\right)}{a+b\cos \left(x\right)}.\frac{a-b\cos \left(y\right)}{\sin \left(y\right)}$

Now $(a-b\cos y)=\frac{a^2-b^2}{a+b\cos x}$

Substituting above,we get

$\frac{\sin \left(x\right)}{a+b\cos \left(x\right)}.\frac{a-b\cos \left(y\right)}{\sin \left(y\right)}$

$=\frac{\sin \left(x\right)}{(a+b\cos (x){)}^2}.\frac{a^2-b^2}{\sin \left(y\right)}$

Hence

$f\left(y\right)=\frac{a^2-b^2}{\sin \left(y\right)}$

Now

$f\left(\frac{\pi }{2}\right)=a^2-b^2$

$=192$.