If - R and 1- i cos-1-2 i cos- is real number- then - will be when I - Set of integersA- 2 n -1 1-2- n - IB- 3 n -2- n - IC- n - n - ID- 2 n - n - I

Z= 1 icosθ1+2icosθ nbsp; On nbsp;rationalization =(1 icosθ1+2icosθ) (1 2icosθ1 2icosθ) nbsp; nbsp; Since, Z is real, hence imaginary part is 0. ⇒ 2cos ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

If θ∈ R and 1...

Question

If $θ \in R$ and $\frac{1 - i cos θ}{1 + 2 i cos θ}$ is real number, then $θ$ will be (when $I$ : Set of integers)

(2 n + 1) \frac{π}{2}, n \in I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 n π}{2}, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n π, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n π, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(2 n + 1) \frac{π}{2}, n \in I$
$Z = \frac{1 - i cos θ}{1 + 2 i cos θ}$
On rationalization
$= (\frac{1 - i cos θ}{1 + 2 i cos θ}) (\frac{1 - 2 i cos θ}{1 - 2 i cos θ})$

Since, $Z$ is real, hence imaginary part is $0$ .
$\Rightarrow \frac{- 2 cos θ - cos θ}{1 + 4 {cos}^{2} θ} = 0$
$\Rightarrow cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, (n \in I)$

Suggest Corrections

Similar questions

The general solution to $s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$ is

$\sqrt{3} + i = (a + i b) (c + i d), t h e n {tan}^{- 1} \frac{b}{a} + {tan}^{- 1} \frac{d}{c}$

has the value

Q. If

A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]

, then A^T + A = I₂, if
(a) θ = n π, n ∈ Z
(b) θ = (2n + 1)

\frac{π}{2}

, n ∈ Z
(c) θ = 2n π +

\frac{π}{3}

, n ∈ Z
(d) none of these

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

Q. The expression

(cos 3 θ + sin 3 θ) + (2 sin 2 θ - 3) (sin θ - cos θ)

is positive for all

θ \in R

Join BYJU'S Learning Program

If θ∈R and 1−icosθ1+2icosθ is real number, then θ will be (when I : Set of integers)

The correct option is A (2n+1)π2,n∈IZ=1−icosθ1+2icosθ On rationalization =(1−icosθ1+2icosθ)(1−2icosθ1−2icosθ) Since, Z is real, hence imaginary part is 0. ⇒−2cosθ−cosθ1+4cos2θ=0 ⇒cosθ=0⇒θ=(2n+1)π2,(n∈I)

If $θ \in R$ and $\frac{1 - i cos θ}{1 + 2 i cos θ}$ is real number, then $θ$ will be (when $I$ : Set of integers)