Trigonometry
Inverse Trigonometric Functions
MCQ (Single Correct Answer)Subjective
1
WB JEE 2021
MCQ (More than One Correct Answer)
+2
-0
If $$\left| {z + i} \right| - \left| {z - 1} \right| = \left| z \right| - 2 = 0$$ for a complex number z, then z is equal to
A
$$\sqrt 2 (1 + i)$$
B
$$\sqrt 2 (1 - i)$$
C
$$\sqrt 2 ( - 1 + i)$$
D
$$\sqrt 2 ( - 1 - i)$$
2
WB JEE 2019
MCQ (More than One Correct Answer)
+2
-0
If $$\theta \in R$$ and $${{1 - i\cos \theta } \over {1 + 2i\cos \theta }}$$ is real number, then $$\theta $$ will be (when I : Set of integers)
A
$$(2n + 1){\pi \over 2},n \in I$$
B
$${{3n\pi } \over 2},n \in I$$
C
$$n\pi ,n \in I$$
D
$$2n\pi ,n \in I$$
3
WB JEE 2017
MCQ (More than One Correct Answer)
+2
-0
The complex number z satisfying the equation | z $$-$$ 1 | = | z + 1 | = 1 is
A
0
B
1 + i
C
$$-$$ 1 + i
D
1 $$-$$ i
4
WB JEE 2016
MCQ (More than One Correct Answer)
+2
-0
If z = sin$$\theta$$ $$-$$ icos$$\theta$$, then for any integer n,
A
$${z^n} + {1 \over {{z^n}}} = 2\cos \left( {{{n\pi } \over 2} - n\theta } \right)$$
B
$${z^n} + {1 \over {{z^n}}} = 2\sin \left( {{{n\pi } \over 2} - n\theta } \right)$$
C
$${z^n} - {1 \over {{z^n}}} = 2i\sin \left( {n\theta - {{n\pi } \over 2}} \right)$$
D
$${z^n} - {1 \over {{z^n}}} = 2i\cos \left( {{{n\pi } \over 2} - n\theta } \right)$$
WB JEE Subjects