Cos-isin-4-icos-sin-5 is equal to

The correct option is D (cos θ +isin θ)^4/(icos θ +sin θ)^5=(cos θ +isin θ)^4/i^5. (cos θ+isin θ)^5 = i(cos 9 θ +i sin 9 θ)=sin 9 θ i cos 9 θ ...

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Trigonometric Ratios of Multiples of an Angle

cosθ+isinθ4/i...

Question

$\frac{(c o s θ + i s i n θ)^{4}}{(i c o s θ + s i n θ)^{5}}$ is equal to

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Solution

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\frac{{(cos θ + i sin θ)}^{4}}{{(sin θ + i cos θ)}^{5}}

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