Question

If $tan\theta =\sqrt{\frac{3}{2}}$, the sum of the infinite series 1 + 2 $(1-cos\theta )+3(1-cos\theta {)}^2+4(1-cos\theta {)}^3+....\infty$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$1+2x+3x^2+4x^3+\cdots +(r+1)x^r+.....to\infty =(1-x{)}^{-2},[let(1-cos\theta )=x]\therefore series=(1-1+cos\theta {)}^{-2}=se{c}^2\theta =(1+ta{n}^2\theta )=1+\frac{3}{2}=\frac{5}{2}$