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Question Number 129181 by liberty last updated on 13/Jan/21

Given tan^2 x−5tan x=3 has the roots are x_1 and x_2 . Find ∣ cos x_1 .cos x_2 ∣ .

$$\:\mathrm{Given}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{5tan}\:\mathrm{x}=\mathrm{3}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{are}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} .\:\mathrm{Find}\:\mid\:\mathrm{cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:\mid\:. \\ $$

Commented by MJS_new last updated on 13/Jan/21

see my solution to question 128995

$$\mathrm{see}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{question}\:\mathrm{128995} \\ $$

Answered by bramlexs22 last updated on 13/Jan/21

Consider the sum and product of the equation tan^2 x−5tan x−3 = 0 { ((tan x_1 +tan x_2 = 5)),((tan x_1 .tan x_2 = −3)) :} we get tan (x_1 +x_2 )=(5/(1−(−3)))=(5/4) and sin (x_1 +x_2 ) = ± (5/( (√(41)))) from tan x_1 +tan x_2 = 5 we get 5cos x_1 .cos x_2 = sin (x_1 +x_2 ) ∣ cos x_1 .cos x_2 ∣ = (1/( (√(41))))

$$\:\mathrm{Consider}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{and}\:\mathrm{product}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{5tan}\:\mathrm{x}−\mathrm{3}\:=\:\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{tan}\:\mathrm{x}_{\mathrm{1}} +\mathrm{tan}\:\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{5}}\\{\mathrm{tan}\:\mathrm{x}_{\mathrm{1}} .\mathrm{tan}\:\mathrm{x}_{\mathrm{2}} \:=\:−\mathrm{3}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{tan}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)=\frac{\mathrm{5}}{\mathrm{1}−\left(−\mathrm{3}\right)}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{and}\:\mathrm{sin}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)\:=\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{41}}} \\ $$$$\mathrm{from}\:\mathrm{tan}\:\mathrm{x}_{\mathrm{1}} +\mathrm{tan}\:\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{5}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\:\mathrm{5cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{sin}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right) \\ $$$$\:\mid\:\mathrm{cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:\mid\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{41}}} \\ $$$$\: \\ $$

Commented by liberty last updated on 13/Jan/21

nice...

$$\mathrm{nice}... \\ $$

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