The trigonometric form of z=(1−i cot 8)3 (where i=−1 ) is

z=(1−i cot8)3 =cosec38(sin8−i cos8)3 =cosec38(cos(π2−8)−i sin(π2−8))3 =cosec38(cos(3π2−24)−i sin(3π2−24))=cosec38.e−i(3π/2−24)=cosec38.ei(24−3π/2)

The trigonometric form of z=(1−i cot 8)3 (where i=−1 ) is

z=(1−i cot8)3 =cosec38(sin8−i cos8)3 =cosec38(cos(π2−8)−i sin(π2−8))3 =cosec38(cos(3π2−24)−i sin(3π2−24))=cosec38.e−i(3π/2−24)=cosec38.ei(24−3π/2)

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The trigonometric form of $z = {(1 - i \cot 8)}^{3}$ (where $i = \sqrt{- 1}$ ) is

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$\cos e c^{3} 8. e^{i (24 - 3 π / 2)}$

$\cos e c^{3} 8. e^{- i (24 - 3 π / 2)}$

$\cos e c^{3} 8. e^{i (36 - π / 2)}$

$\cos e c^{2} 8. e^{- 24 i + π / 2}$

answer is A.

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Detailed Solution

$z = {(1 - i \cot 8)}^{3}$

$\begin{matrix} = \cos e c^{3} 8 {(\sin 8 - i \cos 8)}^{3} \\ = \cos e c^{3} 8 {(\cos (\frac{π}{2} - 8) - i \sin (\frac{π}{2} - 8))}^{3} \\ = \cos e c^{3} 8 (\cos (\frac{3 π}{2} - 24) - i \sin (\frac{3 π}{2} - 24)) \\ = \cos e c^{3} 8. e^{- i (3 π / 2 - 24)} \\ = \cos e c^{3} 8. e^{i (24 - 3 π / 2)} \end{matrix}$