Solve for #a#: #2 tan x - tan 2x +2a=1-tan2x tan^2 x# ?

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Answer 1 · 2017-11-22T17:05:04

# a=1/2.#

We know that, #tan2x=(2tanx)/(1-tan^2x)=(2t)/(1-t^2), where, t=tanx.#

#:. 2tanx-tan2x+2a=1-tan2xtan^2x,#

#rArr 2t-(2t)/(1-t^2)+2a=1-(2t)/(1-t^2)*t^2.#

Multiplying by #(1-t^2),# we get,

#2t(1-t^2)-2t+2a(1-t^2)=(1-t^2)-2t^3.#

#:. cancel(2t-2t^3-2t)+2a(1-t^2)=1-t^2cancel(-2t^3), or,#

# 2a(1-t^2)=(1-t^2).#

# :. 2a=(1-t^2)/(1-t^2)=1, if t^2ne1.#

#:. a=1/2, if t^2ne1.#

Now, #t^2=1 rArr t=tanx=+-1=tan(pmpi/4),#

# rArr x=npi+-pi/4, n in ZZ,# but, then, #tan2x# becomes undefined.

# :. t=tanxnepm1.#

#:. a=1/2.#