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