If 1+ sin θ+ sin 2 θ+⋅s ⋅ upto ∞=2 √3+4, then θ=.............

Question

If 1+ sin θ+ sin 2 θ+⋅s ⋅ upto ∞=2 √3+4, then θ=............. (A) (π/6) (B) (π/4) (C) (π/3) (D) (3π/4)

Tardigrade Admin · Accepted Answer

We have, 1+ sin θ+ sin 2 θ+ ldots ∞=2 √3+4 ⇒ (1/1- sin θ)=2 √3+4 [∵ 1+r+r2+ ldots ∞=(1/1-r)] ⇒ 1- sin θ=(1/2 √3+4) ⇒ sin θ=1-(1/2 √3+4) ⇒ sin θ=(2 √3+4-1/2 √3+4) ⇒ sin θ=(2 √3+3/2 √3+4)=(√3(2+√3)/2(√3+2)) ⇒ sin θ=(√3/2) ∴ θ=60°=(π/3)

Q. If $1 + sin θ + sin^{2} θ + \dots \cdot$ upto $\infty = 23 + 4$ , then $θ =$ .............

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{3 π}{4}$

Q. If 1+sinθ+sin2θ+⋯⋅ upto ∞=23​+4, then θ=.............

6π​

4π​

3π​

43π​

We have, 1+sinθ+sin2θ+…∞=23​+4 ⇒1−sinθ1​=23​+4 [∵1+r+r2+…∞=1−r1​] ⇒1−sinθ=23​+41​ ⇒sinθ=1−23​+41​ ⇒sinθ=23​+423​+4−1​ ⇒sinθ=23​+423​+3​=2(3​+2)3​(2+3​)​ ⇒sinθ=23​​ ∴θ=60∘=3π​

Q. If $1 + sin θ + sin^{2} θ + \dots \cdot$ upto $\infty = 23 + 4$ , then $θ =$ .............

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{3 π}{4}$