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$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$ ની કિંમત શોધો.

જો $\tan A = -\frac{1}{2}$ અને $\tan B = -\frac{1}{3}$ હોય,તો $A + B = $

સાબિત કરો કે $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$

જો $A > B$ અને $\tan A - \tan B = x$ તથા $\cot B - \cot A = y$ હોય,તો $\cot (A - B) = $

$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = $

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