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Show that (tan 30 + x) = 2 tan (60- x) can be written in tan^2 x + (6sqrt3) tan x -5= 0.
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What I did was, (tan 30 + tan x)/1 + tan30tanx = 2 (tan60-tanx)/1-tan60tanx sqrt3/3 + tan x / 1+sqrt3/3 tanx = 2sqrt3 - 2tanx
Is it right so far?
Do you mean: \[ (tan 30 + x) = 2 tan (60- x)\]Is the same as:\[ tan^2 (x) + \sqrt[6]{3}\times tan (x) -5= 0. \]
It's okay. I know where I went wrong alrd.
Ok then. If you still need help, just say so
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