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 ...

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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

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\frac{3 n π}{2}, n \in I

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n π, n \in I

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2 n π, n \in I

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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 $0$ .
$\Rightarrow \frac{- 2 cos θ - cos θ}{1 + 4 {cos}^{2} θ} = 0$
$\Rightarrow cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, (n \in I)$

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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)