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Question Number 128336 by liberty last updated on 06/Jan/21

If ((sec x+tan x)/(sec x−tan x)) = 5 then what is the value of ((3cos 2x+1)/(3cos 2x−1)) ?

$$\mathrm{If}\:\:\frac{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}}{\mathrm{sec}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{5}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{3cos}\:\mathrm{2x}+\mathrm{1}}{\mathrm{3cos}\:\mathrm{2x}−\mathrm{1}}\:?\: \\ $$

Answered by bramlexs22 last updated on 06/Jan/21

⇒ ((sec x+tan x)/(sec x−tan x)) = 5 ⇒((cos x(sec x+tan x))/(cos x(sec x−tan x))) = 5 ⇒((1+sin x)/(1−sin x)) = 5 ; 1+sin x=5−5sin x ⇒6sin x=4 ; sin x = (2/3) then ((3cos 2x+1)/(3cos 2x−1)) = ((3(1−2sin^2 x)+1)/(3(1−2sin^2 x)−1)) ⇒((4−2((4/9)))/(2−2((4/9)))) = ((2−(4/9))/(1−(4/9))) = ((14)/5) .

$$\:\Rightarrow\:\frac{\mathrm{sec}\:{x}+\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}−\mathrm{tan}\:{x}}\:=\:\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}\:=\:\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\:=\:\mathrm{5}\:;\:\mathrm{1}+\mathrm{sin}\:{x}=\mathrm{5}−\mathrm{5sin}\:{x} \\ $$$$\Rightarrow\mathrm{6sin}\:{x}=\mathrm{4}\:;\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${then}\:\frac{\mathrm{3cos}\:\mathrm{2}{x}+\mathrm{1}}{\mathrm{3cos}\:\mathrm{2}{x}−\mathrm{1}}\:=\:\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\right)+\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\right)−\mathrm{1}} \\ $$$$\:\Rightarrow\frac{\mathrm{4}−\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{9}}\right)}{\mathrm{2}−\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{9}}\right)}\:=\:\frac{\mathrm{2}−\frac{\mathrm{4}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}}\:=\:\frac{\mathrm{14}}{\mathrm{5}}\:.\: \\ $$

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