Let \((g \circ f)(x)=\sin x\) and \(f \circ g(x)=(\sin \sqrt{x})^2\) Then,

\(\because g(f(x))=\sin x\) and \(f(g(x))=(\sin \sqrt{x})^2\)
(a) \(f(x)=\sin ^2 x\) and \(g(x)=x\)
Now, \(f(g(x))=f(x)=\sin ^2 x\)
and \(g(f(x))=g\left(\sin ^2 x\right)=\sin ^2 x\)
(b) \(f(x)=\sin \sqrt{x}\) and \(g(x)=\sqrt{x}\)
Now, \(f(g(x))=f(\sqrt{x})=\sin \sqrt{\sqrt{x}}=\sin (x)^4\)
and \(a(f(x))=a(\sin \sqrt{x})=\sqrt{\sin \sqrt{x}}\)
(c) \(f(x)=\sin ^2 x\) and \(g(x)=\sqrt{x}\)
Now, \(f(g(x))=f(\sqrt{x})=\sin ^2 \sqrt{x}\)
and \(g(f(x))=g\left(\sin ^2 x\right)=\sqrt{\sin ^2 x}=|\sin x|\)
(d) \(f(x)=\sin \sqrt{x}\) and \(g(x)=x^2\)
Now, \(f(g(x))=f\left(x^2\right)=\sin |x|\)
and \(g(f(x))=g(\sin \sqrt{x})=(\sin \sqrt{x})^2=\sin ^2 \sqrt{x}\)