(A) Given that $\sin A, \cos A, \tan A$ are in $G.P.$Therefore,the square of the middle term is equal to the product of the extremes:$\cos^2 A = \sin

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(A) Given that $\sin A, \cos A, \tan A$ are in $G.P.$Therefore,the square of the middle term is equal to the product of the extremes:$\cos^2 A = \sin A \cdot \tan A$Substitute $\tan A = \frac{\sin A}{\cos A}$:$\cos^2 A = \sin A \cdot \frac{\sin A}{\cos A}$$\cos^3 A = \sin^2 A$Using the identity $\sin^2 A = 1 - \cos^2 A$:$\cos^3 A = 1 - \cos^2 A$Rearranging the terms:$\cos^3 A + \cos^2 A = 1$

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

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If $\sin A, \cos A$ and $\tan A$ are in $G.P.$,then $\cos^3 A + \cos^2 A$ is equal to

A
$1$
B
$2$
C
$4$
D
None of these

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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If $\sin A, \cos A$ and $\tan A$ are in $G.P.$,then $\cos^3 A + \cos^2 A$ is equal to

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The expression $\frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x}$ is equal to

If $\sin A + \cos A = \sqrt{2}$,then $\cos^2 A = $

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

It is known that $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Then the value of $\frac{\sqrt{3} \sin(\alpha + \beta) - \frac{2}{\cos(\pi/6)} \cos(\alpha + \beta)}{\sin \alpha}$ is:

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