Let S= θ ∈[-π, π]- ± (π/2): sin θ tan θ+ tan θ= sin 2 θ . If T = displaystyle∑θ ∈ S cos 2 θ, then T + n ( S ) is equal

Question

Let S= θ ∈[-π, π]- ± (π/2): sin θ tan θ+ tan θ= sin 2 θ . If T = displaystyle∑θ ∈ S cos 2 θ, then T + n ( S ) is equal (A) 7+√3 (B) 9 (C) 8+√3 (D) 10

Tardigrade Admin · Accepted Answer

sin θ tan θ+ tan θ= sin 2 θ tan θ( sin θ+1)=(2 tan θ/1+ tan 2 θ) tan θ=0 ⇒ θ=-π, 0, π ( sin θ+1)=2 ⋅ cos 2 θ=2(1+ sin θ)(1- sin θ) sin θ=-1 which is not possible sin θ=(1/2) θ=(π/6), (5 π/6) n ( s )=5 T = cos 0+ cos 2 π+ cos 2 π+ cos (π/3)+ cos (5 π/3) T =4 T = n ( s )=9

Q. Let $S = {θ \in [- π, π] - {\pm \frac{π}{2}} : sin θ tan θ + tan θ = sin 2 θ} .$ If $T = θ \in S \sum cos 2 θ$ , then $T + n (S)$ is equal

$7 + 3$

$9$

$8 + 3$

$10$

Q. Let S={θ∈[−π,π]−{±2π​}:sinθtanθ+tanθ=sin2θ}. If T=θ∈S∑​cos2θ, then T+n(S) is equal

7+3​

9

8+3​

10

sinθtanθ+tanθ=sin2θ tanθ(sinθ+1)=1+tan2θ2tanθ​ tanθ=0⇒θ=−π,0,π (sinθ+1)=2⋅cos2θ=2(1+sinθ)(1−sinθ) sinθ=−1 which is not possible sinθ=21​θ=6π​,65π​ n(s)=5 T=cos0+cos2π+cos2π+cos3π​+cos35π​ T=4 T=n(s)=9

Q. Let $S = {θ \in [- π, π] - {\pm \frac{π}{2}} : sin θ tan θ + tan θ = sin 2 θ} .$ If $T = θ \in S \sum cos 2 θ$ , then $T + n (S)$ is equal

$7 + 3$

$9$

$8 + 3$

$10$