If `theta_1a n dtheta_2` are two values lying in `[2,2pi]` for which `t a ntheta=lambda,` then `tan(theta_1)/2tan(theta_2)/2` is equal to 0 (b) `-1` (c) 2 (d) 1

Options

A. 0
B. `-1`
C. 2
D. 1

Answer

B

Explanation

`tan theta=lambda`, we get `(2tan theta//2)/(1-tan^(2)theta/2)=lambda`
or `lambdatan^(2)((theta)/(2))+2tan((theta)/(2))-lambda=0`
`rArr tan((theta_(1))/(2)) tan ((theta_(2))/(2))=-1`