1
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If sin6$$\theta$$ + sin4$$\theta$$ + sin2$$\theta$$ = 0, then general value of $$\theta$$ is
A
$${{n\pi } \over 4},n\pi \pm {\pi \over 3}$$
B
$${{n\pi } \over 4},n\pi \pm {\pi \over 6}$$
C
$${{n\pi } \over 4},2n\pi \pm {\pi \over 3}$$
D
$${{n\pi } \over 4},2n\pi \pm {\pi \over 6}$$
2
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If $$0 \le A \le {\pi \over 4}$$, then $${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$
A
$${\pi \over 4}$$
B
$$\pi$$
C
0
D
$${\pi \over 2}$$
3
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Without changing the direction of the axes, the origin is transferred to the point (2, 3). Then the equation x2 + y2 $$-$$ 4x $$-$$ 6y + 9 = 0 changes to
A
x2 + y2 + 4 = 0
B
x2 + y2 = 4
C
x2 + y2 $$-$$ 8x $$-$$ 12y + 48 = 0
D
x2 + y2 = 9
4
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x $$-$$ 6y + 9sin2$$\alpha$$ + 13cos2$$\alpha$$ = 0 is 2$$\alpha$$. The equation of the locus of the point P is
A
x2 + y2 + 4x + 6y + 9 = 0
B
x2 + y2 $$-$$ 4x + 6y + 9 = 0
C
x2 + y2 $$-$$ 4x $$-$$ 6y + 9 = 0
D
x2 + y2 + 4x $$-$$ 6y + 9 = 0
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