1.

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

Answer» `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`
`tan(theta_1/2)tan(theta_2/2)=((-1+sqrt(1+lambda^2))/lambda)((-1-sqrt(1+lambda^2))/lambda)=-1`.


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