(C) Given $\sin x + \sin^{2} x = 1$. This implies $\sin x = 1 - \sin^{2} x$,so $\sin x = \cos^{2} x$. Squaring both sides,we get $\sin^{2} x = \cos^{4

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(C) Given $\sin x + \sin^{2} x = 1$. This implies $\sin x = 1 - \sin^{2} x$,so $\sin x = \cos^{2} x$. Squaring both sides,we get $\sin^{2} x = \cos^{4} x$. Now,consider the expression $\cos^{8} x + 2 \cos^{6} x + \cos^{4} x$. This can be written as $(\cos^{4} x + \cos^{2} x)^{2}$. Substituting $\cos^{4} x = \sin^{2} x$,we get $(\sin^{2} x + \cos^{2} x)^{2}$. Since $\sin^{2} x + \cos^{2} x = 1$,the expression becomes $(1)^{2} = 1$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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If $\sin x + \sin^{2} x = 1$,then the value of $\cos^{8} x + 2 \cos^{6} x + \cos^{4} x$ is

MHT CET 2020 All MHT CET

A
$3$
B
$2$
C
$1$
D
$4$

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Trigonometrical Ratios, Functions and Identities

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to:

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