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

(d) 1 + 2(1 - cos$\theta$) + 3(1 - cos$\theta$)${}^2$ + 4(1 - cos$\theta$)${}^3$+........ to $\infty$
= (1 - x)${}^{-1}$, [Let (1 - cos$\theta$) = x]
$\therefore$ Series = (1 - 1 + cos$\theta$)${}^{-2}$ = sec${}^2$$\theta$
= (1 + tan${}^2$$\theta$) = 1 + $\frac{3}{2}$ = $\frac{5}{2}$