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Question Number 8105 by 314159 last updated on 30/Sep/16
Prove that ,for any acute angle α   secα cosecα +tanα +cotα ≥4.
$${Prove}\:{that}\:,{for}\:{any}\:{acute}\:{angle}\:\alpha\: \\ $$$${sec}\alpha\:{cosec}\alpha\:+{tan}\alpha\:+{cot}\alpha\:\geqslant\mathrm{4}. \\ $$
Commented by Yozzia last updated on 30/Sep/16
f(x)=(1/(cosxsinx))+((sinx)/(cosx))+((cosx)/(sinx))  f(x)=(1/(cosxsinx))+(1/(sinxcosx))  f(x)=(2/(cossinx))=(4/(sin2x))=4cosec2x  Since 0<α<90°⇒0<2α<180°  ⇒0<sin2x≤1⇒cosec2x≥1  ⇒4cosec2x≥4⇒f(x)≥4  ∴ secxcosecx+tanx+cotx≥4 ∀x∈(0,90°).
$${f}\left({x}\right)=\frac{\mathrm{1}}{{cosxsinx}}+\frac{{sinx}}{{cosx}}+\frac{{cosx}}{{sinx}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{cosxsinx}}+\frac{\mathrm{1}}{{sinxcosx}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{{cossinx}}=\frac{\mathrm{4}}{{sin}\mathrm{2}{x}}=\mathrm{4}{cosec}\mathrm{2}{x} \\ $$$${Since}\:\mathrm{0}<\alpha<\mathrm{90}°\Rightarrow\mathrm{0}<\mathrm{2}\alpha<\mathrm{180}° \\ $$$$\Rightarrow\mathrm{0}<{sin}\mathrm{2}{x}\leqslant\mathrm{1}\Rightarrow{cosec}\mathrm{2}{x}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{cosec}\mathrm{2}{x}\geqslant\mathrm{4}\Rightarrow{f}\left({x}\right)\geqslant\mathrm{4} \\ $$$$\therefore\:{secxcosecx}+{tanx}+{cotx}\geqslant\mathrm{4}\:\forall{x}\in\left(\mathrm{0},\mathrm{90}°\right). \\ $$
Answered by prakash jain last updated on 02/Oct/16
answer in coments
$$\mathrm{answer}\:\mathrm{in}\:\mathrm{coments} \\ $$

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