Verified Solution • 2025–26 Session moderate

QJEEM20Maf09e: JEE Main 2020 Mathematics - Class 11 JEEM20Maf09e Solution: If , [0, 2 ], is a real number, then an argument of sin + icos…

If 3+isin⁡θ4−icos⁡θ{{3 + i\sin \theta } \over {4 - i\cos \theta }} , θ\theta ∈\in [0, 2 θ\theta ], is a real number, then an argument of
sin θ\theta + icos θ\theta is :

Multiple Choice Options

  • A π−tan⁡−1(34)\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)
  • B −tan⁡−1(34)- {\tan ^{ - 1}}\left( {{3 \over 4}} \right)
  • C tan⁡−1(43){\tan ^{ - 1}}\left( {{4 \over 3}} \right)
  • D π−tan⁡−1(43)\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)

Step-by-Step Verified Solution

Let z = 3+isin⁡θ4−icos⁡θ{{3 + i\sin \theta } \over {4 - i\cos \theta }}

= 3+isin⁡θ4−icos⁡θ×(4+icos⁡θ)(4+icos⁡θ){{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}

= (12−sin⁡θcos⁡θ)+i(4sin⁡θ+3cos⁡θ)16+cos⁡2θ{{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}

As Z is purely real

∴\therefore 4sin θ\theta + 3cos θ\theta = 0

⇒\Rightarrow tan θ\theta = −34- {3 \over 4}

∴\therefore θ\theta lies in the 2nd quadrant then

arg(sinθ + icosθ) = π\pi + tan⁡−1(cos⁡θsin⁡θ){\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)

= π−tan⁡−1(43)\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)

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