(B) Given $A = \sin^8\theta + \cos^{14}\theta$. Since $0 \le \sin^2\theta \le 1$ and $0 \le \cos^2\theta \le 1$,we have $\sin^8\theta \le \sin^2\theta

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(B) Given $A = \sin^8\theta + \cos^{14}\theta$. Since $0 \le \sin^2\theta \le 1$ and $0 \le \cos^2\theta \le 1$,we have $\sin^8\theta \le \sin^2\theta$ and $\cos^{14}\theta \le \cos^2\theta$. Adding these inequalities,we get $A = \sin^8\theta + \cos^{14}\theta \le \sin^2\theta + \cos^2\theta = 1$. Also,since $\sin^2\theta$ and $\cos^2\theta$ cannot be zero simultaneously,$A > 0$. Thus,$0 < A \le 1$.

Class 11 Mathematics Trigonometric Functions Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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If $A = \sin^8\theta + \cos^{14}\theta$,then for all real values of $\theta$:

A
$A \ge 1$
B
$0 < A \le 1$
C
$1 < 2A \le 3$
D
None of these

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Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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If $A = \sin^8\theta + \cos^{14}\theta$,then for all real values of $\theta$:

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