If sin x+sin ^2 x=1,where x inleft(0, frac{pi | vedclass

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(C) Given $\sin x+\sin ^2 x=1$,we have $\sin x=1-\sin ^2 x=\cos ^2 x$.Since $\sin x=\cos ^2 x$,it follows that $	an x = \frac{\sin x}{\cos x} = \frac

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$2$

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(C) Given $\sin x+\sin ^2 x=1$,we have $\sin x=1-\sin ^2 x=\cos ^2 x$.Since $\sin x=\cos ^2 x$,it follows that $	an x = \frac{\sin x}{\cos x} = \frac{\cos ^2 x}{\cos x} = \cos x$.Substituting $	an x = \cos x$ into the expression,we get:$(\cos ^{12} x+\cos ^{12} x)+3(\cos ^{10} x+\cos ^{10} x+\cos ^8 x+\cos ^8 x)+(\cos ^6 x+\cos ^6 x)$$= 2\cos ^{12} x + 6\cos ^{10} x + 6\cos ^8 x + 2\cos ^6 x$$= 2(\cos ^{12} x + 3\cos ^{10} x + 3\cos ^8 x + \cos ^6 x)$$= 2(\cos ^4 x + \cos ^2 x)^3$Since $\cos ^4 x + \cos ^2 x = \sin ^2 x + \sin x = 1$,the expression becomes $2(1)^3 = 2$.

If $\sin x+\sin ^2 x=1$,where $x \in\left(0, \frac{\pi}{2}\right)$,then the value of $(\cos ^{12} x+\tan ^{12} x)+3(\cos ^{10} x+\tan ^{10} x+\cos ^8 x+\tan ^8 x)+(\cos ^6 x+\tan ^6 x)$ is equal to

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