If `sinthetaa n d-costheta` are the roots of the equation `a x^2-b x-c=0` , where `a , ba n dc` are the sides of a triangle ABC, then `cosB` is equal to `1-c/(2a)` (b) `1-c/a` `1+c/(c a)` (d) `1+c/(3a)`

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)`