1.

Prove that `(tan3x)/(tanx)`never lies between `1/3a n d3`.

Answer» `(tan3x)/tanx = (3tan - tan^3x)/(tanx(1-3tan^2x))`
`=(3-tan^2x)/(1-3tan^2x)`
Now, let `(3-tan^2x)/(1-3tan^2x) = a`
`=>3-tan^2x = a-3atan^2x`
`=>tan^2x(3a-1) = a-3`
`=>tan^2x = (a-3)/(3a-1)`
It means, `(a-3)/(3a-1) gt 0.`So, `a-3 gt 0 and 3a-1 lt 0`
`=> a gt 3 and a lt 1/3`
So, `(tan3x)/tanx` never lies between `1/3` and `3.`


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