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If $-i$ and $\alpha$ are the roots of the equation $i z^2-2(i+1) z+$ $(2-i)=0, \tan \theta=\frac{-1}{2}$ and $\theta \in 4^{\text {th }}$ quadrant, then $5^3 \cos 6 \theta=$
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–117
We have $\tan \theta=\frac{-1}{2} \Rightarrow \tan 3 \theta=\frac{-11}{2}$
Now $5^3 \cos 6 \theta=125\left(\frac{1-\tan ^2(3 \theta)}{1+\tan ^2(3 \theta)}\right)$ $=125 \times \frac{-117}{125}=-117$
Now $5^3 \cos 6 \theta=125\left(\frac{1-\tan ^2(3 \theta)}{1+\tan ^2(3 \theta)}\right)$ $=125 \times \frac{-117}{125}=-117$
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