jee-main 2022 Q64

jee-main · India · session2_27jul_shift2 Standard trigonometric equations Solve trigonometric equation for solutions in an interval

Let $S = \left\{ \theta \in \left( 0 , \frac { \pi } { 2 } \right) : \sum _ { m = 1 } ^ { 9 } \sec \left( \theta + ( m - 1 ) \frac { \pi } { 6 } \right) \sec \left( \theta + \frac { m \pi } { 6 } \right) = - \frac { 8 } { \sqrt { 3 } } \right\}$. Then
(1) $\mathrm { S } = \left\{ \frac { \pi } { 12 } \right\}$
(2) $S = \left\{ \frac { 2 \pi } { 3 } \right\}$
(3) $\sum _ { \theta \in S } \theta = \frac { \pi } { 2 }$
(4) $\sum _ { \theta \in S } \theta = \frac { 3 \pi } { 4 }$

Let $S = \left\{ \theta \in \left( 0 , \frac { \pi } { 2 } \right) : \sum _ { m = 1 } ^ { 9 } \sec \left( \theta + ( m - 1 ) \frac { \pi } { 6 } \right) \sec \left( \theta + \frac { m \pi } { 6 } \right) = - \frac { 8 } { \sqrt { 3 } } \right\}$. Then\\
(1) $\mathrm { S } = \left\{ \frac { \pi } { 12 } \right\}$\\
(2) $S = \left\{ \frac { 2 \pi } { 3 } \right\}$\\
(3) $\sum _ { \theta \in S } \theta = \frac { \pi } { 2 }$\\
(4) $\sum _ { \theta \in S } \theta = \frac { 3 \pi } { 4 }$

Paper Questions

Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86 Q87 Q88 Q89 Q90