$Minimum value of $\mathbf{\frac{\tan\left(x+\frac{\pi}{6}\right)}{\tan x}}$
66You should specify some conditions on x otherwise there is no minimum.
341The range is
\left( -\infty, \frac{1}{3}\right) \bigcup{}\left(3, \infty \right)
341Let
s = \frac{\tan \left(x+\frac{\pi}{6} \right)} {\tan x}
\frac{s+1}{s-1} = \frac{\tan \left(x+\frac{\pi}{6} \right)+\tan x} {\tan \left(x+\frac{\pi}{6} \right)-\tan x} = \frac{\sin \left(2x+\frac{\pi}{6} \right)}{\sin \frac{\pi}{6}}=2 \sin \left(2x+\frac{\pi}{6} \right)
So we have
-2 \le \frac{s+1}{s-1} \le 2
If s≥1, we obtain that s≥3
And if s≤1, we obtain that s≤1/3 and thus the range of the given expression is obtained
3awesome sir.
i got the same question in my test today.
i did it by this,
let y = tan(x+30)/tanx
then, y = [tanx + tan30]/[(1-tanx*tan30)*tanx]
now here we put tan30 = √3 and make a quadratic in tanx.
the roots will be real so D>0.
and we get our answer.
3i made a mistake there.
it should have been D≥0 not D>0.
:-)
and i put tan30 as √3 and not 1/√3.
