If -i and α are the roots of the equation i z^2-2(i+1) z+ (2-i)=0, tan θ=(-1)/2 and θ ∈ 4^th quadrant, then 5^3 cos 6 θ=

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MARKS Content Team · Accepted Answer

Correct answer: (A) –117. We have tan θ=(-1)/2 ⇒ tan 3 θ=(-11)/2 Now 5^3 cos 6 θ=125 ((1-tan^2(3 θ))/(1+tan^2(3 θ)) ) =125 × (-117)/125=-117

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 $5^3 \cos 6 \theta=$