We have #ain(pi,(3pi)/2)# and #cosa=-2/3#. How to find #tan(2a)#?

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Answer 1 · 2017-06-26T17:01:26

# tan2a=-4sqrt5.#

We will use, #tan2a=tan(a+a)=(tana+tana)/(1-tana*tana), i.e., #

#tan2a=(2tana)/(1-tan^2a).#

Given that, #cosa=-2/3, a in (pi,3pi/2),# we have,

#seca=-3/2 rArr sec^2a-1=tan^2a = 9/4-1=5/4.#

#because, a in (pi,3pi/2), :., tana=+sqrt5/2.#

#:. tan2a=(2*sqrt5/2)/(1-5/4)=sqrt5/(-1/4),#

# rArr tan2a=-4sqrt5.#