tani[loge(a+iba−ib)] is equal to

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i [ _{e} ( {a-ib}{a+ib} ) ] is equal to

Question

$tan i [lo g_{e} (\frac{a - ib}{a + ib})]$ is equal to

\frac{a ^{2} - b ^{2}}{2 ab}

\frac{2 ab}{a ^{2} + b ^{2}}

\frac{2 ab}{a ^{2} - b ^{2}}

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tan i [lo g_{e} (\frac{a - ib}{a + ib})]

is equal to

8 mins ago

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8 mins ago

tan i [lo g_{e} (\frac{cos θ - i sin θ}{cos θ + i sin θ})]

By putting

a = r cos θ

and

b = r sin θ

= tan [i lo g_{e} (cos 2 θ - i sin 2 θ)]

= tan [i lo g_{e} e^{- 2 i θ}]

= tan [i (- 2 i θ)]

= tan 2 θ

= \frac{2 tan θ}{1 - tan ^{2} θ} = \frac{2 ab}{a ^{2} - b ^{2}}

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Question Text	$tan i [lo g_{e} (\frac{a - ib}{a + ib})]$ is equal to
Answer Type	Text solution:1
Upvotes	150