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 \(5^3 \cos 6 \theta=\)
- A –117
- B -44
- C 117
- D 44
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 \(5^3 \cos 6 \theta=125\left(\frac{1-\tan ^2(3 \theta)}{1+\tan ^2(3 \theta)}\right)\) \(=125 \times \frac{-117}{125}=-117\)
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