3-2 i sin-1-2 i sin- will be purely imaginary- if -Where n is an integer- -IIT 1976- Pb- CET 2003-A- 2 n -3B- n -3C- n -3D- None of these

3 + 2i sin θ/1 2i sin θ will be purely imaginary, if the real part vanishes, i.e., 3 4 sin^2 θ/1 + 4 sin^2 θ = 0 ⇒ 3 4 sin^2θ = 0 (only if θ be real ...

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Standard XII

Mathematics

Purely Imaginary

3+2 i sinθ/1-...

Question

$\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}$ will be purely imaginary, if $θ$ =

[Where n is an integer]
[IIT 1976; Pb. CET 2003]

2 n π \pm \frac{π}{3}

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n π + \frac{π}{3}

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n π \pm \frac{π}{3}

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None of these

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Solution

The correct option is C $n π \pm \frac{π}{3}$
$\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}$ will be purely imaginary, if the real part vanishes, i.e., $\frac{3 - 4 s i n^{2} θ}{1 + 4 s i n^{2} θ} = 0$
$\Rightarrow 3 - 4 s i n^{2} θ = 0$ (only if $θ$ be real)
$\Rightarrow s i n θ \pm \frac{\sqrt{3}}{2} = s i n (\pm \frac{π}{3})$
$\Rightarrow θ = n π + (- 1)^{n} (\pm \frac{π}{3}) = n π \pm \frac{π}{3}$

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Similar questions

\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}

will be purely imaginary, if

θ

=

[Where n is an integer]

[IIT 1976; Pb. CET 2003]

Q. The general solution of

t a n^{2} θ = 3 i s (n ϵ Z)

L e t A = {θ \in (- \frac{π}{2}, π) : \frac{3 + 2 i sin θ}{1 - 2 i sin θ} is purely imaginary}

. The sum of all the elements in

A

is :

Q. The set of values of

x

satisfying the equation

1 + cos 3 x = 2 cos 2 x

can be
(where

n \in Z

)

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

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3+2i sin θ1−2i sin θ will be purely imaginary, if θ = [Where n is an integer] [IIT 1976; Pb. CET 2003]

The correct option is C nπ±π33+2i sin θ1−2i sin θ will be purely imaginary, if the real part vanishes, i.e., 3−4 sin2 θ1+4 sin2 θ=0 ⇒ 3−4sin2θ=0 (only if θ be real) ⇒ sin θ±√32=sin(±π3) ⇒ θ=nπ+(−1)n(± π3)=nπ±π3

$\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}$ will be purely imaginary, if $θ$ =

[Where n is an integer]
[IIT 1976; Pb. CET 2003]