Question

The sum of the series sin$\theta$.sec3$\theta$ + sin3$\theta$.$sec{3}^2$$\theta$ + $sin{3}^2$$\theta$.sec$3^3$$\theta$ + ...........n terms is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

S = $\sum _{r-1}^n$ $sin{3}^{r-1}$$\theta$.$sec{3}^r$$\theta$

= $\sum _{r-1}^n$ $\frac{sin{3}^{r-1}\theta }{cos{3}^r\theta }$

=$\sum _{r-1}^n$ $\frac{2cos{3}^{r-1}\theta .sin{3}^{r-1}\theta }{2cos{3}^{r-1}\theta .cos{3}^r\theta }$ = $\frac{1}{2}$$\sum _{r-1}^n$ $\frac{sin\left({2.3}^{r-1}\theta \right)}{cos{3}^{r-1}\theta .cos{3}^r\theta }$

= $\frac{1}{2}$ $\sum _{r-1}^n$$\frac{sin{3}^r\theta .cos{3}^r\theta -cos{3}^r\theta .sin3r^{-1}\theta }{cos{3}^{r-1}\theta .cos{3}^r\theta }$

= $\frac{1}{2}$ $\sum _{r-1}^n$$(tan{3}^r\theta -tan{3}^{r-1}\theta )$

= $\frac{1}{2}$ $\left[\right(tan3\theta -tan\theta )+(tan{3}^2\theta -tan3\theta )+.............+(tan{3}^n\theta -tan{3}^{n-1}\theta \left)\right]$

= $\frac{1}{2}$ $[tan{3}^n\theta -tan\theta ]$