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

Explanation

`tan theta=lambda`

`tan2A=(2tanA)/(1-tan^2A)`

`(2tan(theta/2))/(1-tan^2(theta/2))=lambda`

`lambdatan^2(theta/2)+2tan(theta/2)-lambda=0`

`tan(theta/2)=(-2pmsqrt(4+4lambda^2))/(2lambda)`

`=(-1pmsqrt(1+lambda^2))/lambda`

`tan(theta_1/2)=(-1+sqrt(1+lambda^2))/lambda`

`tan(theta_2/2)=(1--sqrt(1+lambda^2))/lambda`

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