JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 12)
If $${{3 + i\sin \theta } \over {4 - i\cos \theta }}$$, $$\theta $$ $$ \in $$ [0, 2$$\theta $$], is a real number, then an argument of
sin$$\theta $$ + icos$$\theta $$ is :
sin$$\theta $$ + icos$$\theta $$ is :
$$\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
$$ - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
$${\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
$$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
Explanation
Let z = $${{3 + i\sin \theta } \over {4 - i\cos \theta }}$$
= $${{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}$$
= $${{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}$$
As Z is purely real
$$ \therefore $$ 4sin$$\theta $$ + 3cos$$\theta $$ = 0
$$ \Rightarrow $$ tan $$\theta $$ = $$ - {3 \over 4}$$
$$ \therefore $$ $$\theta $$ lies in the 2nd quadrant then
arg(sinθ + icosθ) = $$\pi $$ + $${\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)$$
= $$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
= $${{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}$$
= $${{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}$$
As Z is purely real
$$ \therefore $$ 4sin$$\theta $$ + 3cos$$\theta $$ = 0
$$ \Rightarrow $$ tan $$\theta $$ = $$ - {3 \over 4}$$
$$ \therefore $$ $$\theta $$ lies in the 2nd quadrant then
arg(sinθ + icosθ) = $$\pi $$ + $${\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)$$
= $$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
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