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TS EAMCET · Maths · Complex Number

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 \(5^3 \cos 6 \theta=\)

  1. A –117
  2. B -44
  3. C 117
  4. D 44
Verified Solution

Answer & Solution

Correct Answer

(A) –117

Step-by-step Solution

Detailed explanation

We have \(\tan \theta=\frac{-1}{2} \Rightarrow \tan 3 \theta=\frac{-11}{2}\) Now \(5^3 \cos 6 \theta=125\left(\frac{1-\tan ^2(3 \theta)}{1+\tan ^2(3 \theta)}\right)\) \(=125 \times \frac{-117}{125}=-117\)