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if-tan-2-x-sec-x-a-1-has-at-least-one-solution-then-find-the-complete-set-of-values-of-a-




Question Number 146868 by gsk2684 last updated on 16/Jul/21
if tan^2 x+sec x=a+1 has at least   one solution then find the complete set  of values of  ′a′?
$${if}\:\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{sec}\:{x}={a}+\mathrm{1}\:{has}\:{at}\:{least}\: \\ $$$${one}\:{solution}\:{then}\:{find}\:{the}\:{complete}\:{set} \\ $$$${of}\:{values}\:{of}\:\:'{a}'? \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jul/21
tan^2 x+secx = a+1  (1)  1+tan^2 x+secx = a+2  sec^2 x+secx−(a+2) = 0  Δ = 1^2 −4×1(−(a+2) = 4a+9  (1) has at least one solution if Δ≥0  ⇒ a∈[−(9/4);+∞[
$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{sec}{x}\:=\:{a}+\mathrm{1}\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{sec}{x}\:=\:{a}+\mathrm{2} \\ $$$$\mathrm{sec}^{\mathrm{2}} {x}+\mathrm{sec}{x}−\left({a}+\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\Delta\:=\:\mathrm{1}^{\mathrm{2}} −\mathrm{4}×\mathrm{1}\left(−\left({a}+\mathrm{2}\right)\:=\:\mathrm{4}{a}+\mathrm{9}\right. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{if}\:\Delta\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{a}\in\left[−\frac{\mathrm{9}}{\mathrm{4}};+\infty\left[\right.\right. \\ $$
Commented by gsk2684 last updated on 16/Jul/21
thank you for your concern sir.  please verify the bounds of sec x  i.e. sec x ≤ −1 or sec x ≥ 1
$${thank}\:{you}\:{for}\:{your}\:{concern}\:{sir}. \\ $$$${please}\:{verify}\:{the}\:{bounds}\:{of}\:\mathrm{sec}\:{x} \\ $$$${i}.{e}.\:\mathrm{sec}\:{x}\:\leqslant\:−\mathrm{1}\:{or}\:\mathrm{sec}\:{x}\:\geqslant\:\mathrm{1} \\ $$

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