Question #170b1

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Answer 1 · 2017-12-27T10:36:00

# -1#.

Recall that, #tan2x=(2tanx)/(1-tan^2x)#.

#:. tantheta={2tan(theta/2)}/{1-tan^2(theta/2)}#.

Hence, # tantheta=lambda,#

#rArr {2tan(theta/2)}/{1-tan^2(theta/2)}=lambda#.

Letting, #tan(theta/2)=t,# we get,

#(2t)/(1-t^2)=lambda, i.e., lambdat^2+2t-lambda=0............(star)#.

Now, we know from Algebra that if #alpha and beta# are the

roots of the quadr. eqn. #at^2+bt+c=0,# then,

#alpha+beta=-b/a, and, alpha*beta=c/a#.

#"But, "t_1=tan(theta_1/2) and t_2=tan(theta_2/2)# are the roots

of quadratic eqn. #(star)#.

#:. t_1*t_2=tan(theta_1/2)*tan(theta_2/2)=(-lambda)/(lambda)=-1#.