Let S = θ ∈[0,2 π]: 82 sin 2 θ+82 cos 2 θ=16 . Then n ( S )+ displaystyle∑θ ∈ S ( sec ((π/4)+2 θ) operatornamecosec((π/4)+2 θ)) is equal to:

Question

Let S = θ ∈[0,2 π]: 82 sin 2 θ+82 cos 2 θ=16 . Then n ( S )+ displaystyle∑θ ∈ S ( sec ((π/4)+2 θ) operatornamecosec((π/4)+2 θ)) is equal to: (A) 0 (B) -2 (C) -4 (D) 12

Tardigrade Admin · Accepted Answer

82 sin 2 θ+82-2 sin 2 θ=16 y +(64/ y )=16 ⇒ y =8 ⇒ sin 2 θ=1 / 2 n ( S )+ displaystyle∑θ ∈ S (1/ cos (π / 4+2 θ) sin (π / 4+2 θ)) =4+(-2) × 4=-4

Q. Let
$S = {θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then
$n (S) + θ \in S \sum (sec (\frac{π}{4} + 2 θ) cosec (\frac{π}{4} + 2 θ))$
is equal to:

0

$- 2$

$- 4$

12

$8^{2 s i n^{2} θ} + 8^{2 - 2 s i n^{2} θ} = 16$
$y + \frac{64}{y} = 16$
$\Rightarrow y = 8$
$\Rightarrow sin^{2} θ = 1/2$
$n (S) + θ \in S \sum \frac{1}{cos ( π /4 + 2 θ ) sin ( π /4 + 2 θ )}$
$= 4 + (- 2) \times 4 = - 4$

Q. Let S={θ∈[0,2π]:82sin2θ+82cos2θ=16}. Then n(S)+θ∈S∑​(sec(4π​+2θ)cosec(4π​+2θ)) is equal to:

0

−2

−4

12

82sin2θ+82−2sin2θ=16 y+y64​=16 ⇒y=8 ⇒sin2θ=1/2 n(S)+θ∈S∑​cos(π/4+2θ)sin(π/4+2θ)1​ =4+(−2)×4=−4

Q. Let
$S = {θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then
$n (S) + θ \in S \sum (sec (\frac{π}{4} + 2 θ) cosec (\frac{π}{4} + 2 θ))$
is equal to:

$- 2$

$- 4$

$8^{2 s i n^{2} θ} + 8^{2 - 2 s i n^{2} θ} = 16$
$y + \frac{64}{y} = 16$
$\Rightarrow y = 8$
$\Rightarrow sin^{2} θ = 1/2$
$n (S) + θ \in S \sum \frac{1}{cos ( π /4 + 2 θ ) sin ( π /4 + 2 θ )}$
$= 4 + (- 2) \times 4 = - 4$