If (3 + isinθ)/(4 - icosθ), θ ∈ [0 ,2 π], is a real number, then an argument of sinθ + icosθ is

Correct answer: A. π - ^ - 1 ((4)/(3)) z = ((3 + isinθ))/((4 - icosθ)) × ((4 + icosθ))/((4 + icosθ)) Im ( z ) = 3cosθ + 4sinθ 16 + ^ 2 θ As z is purely real ⇒ 3cosθ + 4sinθ = 0 ⇒ tanθ = - (3)/(4) Arg(sinθ + icosθ) = π + ^ - 1 ((cosθ)/(sinθ)) = π + ^ - 1 (- (4)/(3)) = π - ^ - 1 ((4)/(3))

JEE Main 20207 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Complex Number

If (3 + isinθ)/(4 - icosθ), θ ∈ [0 ,2 π], is a real number, then an...

\dfrac{3 + isin\theta}{4 - icos\theta}, \theta \in \left[0 ,2 \pi\right],

is a real number, then an argument of

sin\theta + icos\theta

Options

A $\pi - \tan^{- 1}\left(\dfrac{4}{3}\right)$
B $\pi - \tan^{- 1}\left(\dfrac{3}{4}\right)$
C $- \tan^{- 1}\left(\dfrac{3}{4}\right)$
D $\tan^{- 1}\left(\dfrac{4}{3}\right)$

Correct answer

A. $\pi - \tan^{- 1}\left(\dfrac{4}{3}\right)$

Step-by-step solution

z = \dfrac{\left(3 + isin\theta\right)}{\left(4 - icos\theta\right)} \times \dfrac{\left(4 + icos\theta\right)}{\left(4 + icos\theta\right)} $</p><p>$ Im ( z ) = \dfrac{3cos\theta + 4sin\theta}{16 + \cos^{2}\theta} $<br />As$ z $is purely real$ ⇒ 3cos\theta + 4sin\theta = 0 ⇒ tan\theta = - \dfrac{3}{4} $<br />Arg\left(sin\theta + icos\theta\right) = \pi + \tan^{- 1}\left(\dfrac{cos\theta}{sin\theta}\right) = \pi + \tan^{- 1}\left(- \dfrac{4}{3}\right) = \pi - \tan^{- 1}\left(\dfrac{4}{3}\right)$

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