If `sin theta and -cos theta` are the roots of the equation `ax^(2) - bx - c = 0`, where a, b, and c are the sides of a triangle ABC, then `cos B` is equal to

Options

A. `1 - (c)/(2a)`
B. `1 - (c)/(a)`
C. `1 + (c)/(2a)`
D. `1 + (c)/(3a)`

Answer

C

Explanation

Here, `sin theta - cos theta = (b)/(a) and sin theta cos theta = (c)/(a)`
or `1 - 2 sin theta cos theta = (b^(2))/(a^(2))`
or `1 -(2c)/(a) = (b^(2))/(a^(2))`
or `a^(2) - b^(2) = 2ac`
Hence, `cos B = (a^(2) + c^(2) -b^(2))/(2ac)`
`= (2ac + c^(2))/(2ac) = 1 + (c)/(2a)`