TanA + Tan(60°+A) + Tan(120°+A)
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Answered by
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tan A + root3 + tanA /1-root3tanA + tan 120 + tanA / 1- tan120tanA
tan(180-60) = -tan60 = -root3
tanA + [root3 + tana / 1-root3tanA ]+ [(-root3) +tanA / 1+root3tanA ]
on taking the l.c.m we get
tanA(1- 3tan 2A) +(root3tanA +1)(root3 + tanA) +(tanA -root3)(1-root3tanA )/1-3tan2A
after solving brackets we get
9tanA - 3tan3A /1-3tan2A
3(3tanA - tan3A)/1 -3tan2 A
i.e 3tan3A
tan(180-60) = -tan60 = -root3
tanA + [root3 + tana / 1-root3tanA ]+ [(-root3) +tanA / 1+root3tanA ]
on taking the l.c.m we get
tanA(1- 3tan 2A) +(root3tanA +1)(root3 + tanA) +(tanA -root3)(1-root3tanA )/1-3tan2A
after solving brackets we get
9tanA - 3tan3A /1-3tan2A
3(3tanA - tan3A)/1 -3tan2 A
i.e 3tan3A
bunny3062c:
last step verify once
posted it again
answrr
do theanswr i posted again
your answer is correct sorry
iam very happy because you forgived me
Answered by
0
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SOLUTION
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Apply tan(A+B) formula i.e tanA+tanB /1 -tanAtanB
tan A + root3 + tanA /1-root3tanA + tan 120 + tanA / 1- tan120tanA
tan(180-60) = -tan60 = -root3
tanA + [root3 + tana / 1-root3tanA ]+ [(-root3) +tanA / 1+root3tanA ]
on taking the l.c.m we get
tanA(1- 3tan 2A) +(root3tanA +1)(root3 + tanA) +(tanA -root3)(1-root3tanA )/1-3tan2A
after solving brackets we get
9tanA - 3tan3A /1-3tan2A
3(3tanA - tan3A)/1 -3tan2 A
i.e 3tan3A
SOLUTION
======{{{
Apply tan(A+B) formula i.e tanA+tanB /1 -tanAtanB
tan A + root3 + tanA /1-root3tanA + tan 120 + tanA / 1- tan120tanA
tan(180-60) = -tan60 = -root3
tanA + [root3 + tana / 1-root3tanA ]+ [(-root3) +tanA / 1+root3tanA ]
on taking the l.c.m we get
tanA(1- 3tan 2A) +(root3tanA +1)(root3 + tanA) +(tanA -root3)(1-root3tanA )/1-3tan2A
after solving brackets we get
9tanA - 3tan3A /1-3tan2A
3(3tanA - tan3A)/1 -3tan2 A
i.e 3tan3A
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