Sum of roots of the equation sin x `xx` tan x – sin x + tanx – 1 = 0 over [0, `2pi`] is :
| A. |
`1/2` |
| B. |
`5/2` |
| C. |
3 |
| D. |
`(3pi)/2` |
 |
sin x tan x – sin x + tan x – 1 = 0
tan x (sin x + 1) – (sin x + 1) = 0
(sin x + 1)(tanx – 1) = 0
sin x = – 1 or tan x = 1
`x=(3pi)/2`(rejected) or `x={pi/4 ,(5pi)/4}`
Therefore, the sum of solutions= `pi/4+(5pi)/4=(3pi)/2`