The question seems to be wrong because if we take \(\sin ^{2} x=1-x\), then \(x\) should be around \(30^{\circ}-40^{\circ}\), then none of the choices will be correct. \[\] Question should be \(\sin ^{2} x=1-\sin x\) \[\] \(\Rightarrow 1-\cos ^{2} x=1-\sin x \Rightarrow \cos ^{2} x=\sin x\) \[\] Squaring it again we get the answer \[\] \(\cos ^{4} x=1-\cos ^{2} x \Rightarrow \cos ^{4} x+\cos ^{2} x=1\). \[\] Choice (B)