(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$Factor out $2$: $2(\cos^2 \theta - \sin^2 \theta) = 1$Using the identity $\cos 2\theta = \cos

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(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$Factor out $2$: $2(\cos^2 \theta - \sin^2 \theta) = 1$Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we get:$2\cos 2\theta = 1$Divide by $2$: $\cos 2\theta = \frac{1}{2}$Since $\cos 60^\circ = \frac{1}{2}$,we have:$2\theta = 60^\circ$Therefore,$\theta = 30^\circ$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Trigonometrical ratios of multiple and sub-multiple angles

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If $2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots ^\circ$.

A
$15$
B
$30$
C
$45$
D
$60$

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Trigonometrical Ratios, Functions and Identities

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Trigonometrical ratios of multiple and sub-multiple angles

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If $2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots ^\circ$.

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If $\theta = 3\alpha$ and $\sin \theta = \frac{a}{\sqrt{a^2 + b^2}}$,then the value of the expression $a \csc \alpha - b \sec \alpha$ is:

If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$,find the values of $\sin \frac{x}{2}$,$\cos \frac{x}{2}$,and $\tan \frac{x}{2}$.

The value of $\cos(60^{\circ} - A) \cdot \cos A \cdot \cos(60^{\circ} + A)$ is

If $90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2} =$

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