If-1-sin-x-cos-x-1-sin-x-cos-x-1-4-then-find-the-value-of-tan-x-1-cos-x-

Question Number 161931 by mnjuly1970 last updated on 24/Dec/21

$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}{f}\:\:\:\:\frac{\:\mathrm{1}−{sin}\left({x}\right)−{cos}\left({x}\right)}{\mathrm{1}+{sin}\left({x}\right)−{cos}\left({x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:{then}\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{tan}\left({x}\right)\:+\:\frac{\mathrm{1}}{{cos}\left({x}\right)}\:=? \\ $$$$\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by cortano last updated on 24/Dec/21

((2sin^2 ((x/2))−2sin ((x/2))cos ((x/2)))/(2sin^2 ((x/2))+2sin ((x/2))cos ((x/2)))) =(1/4) ((sin (x/2)−cos (x/2))/(sin (x/2)+cos (x/2))) =(1/4) ((1−sin x)/(cos x)) = (1/4) sec x−tan x=(1/4) . let tan x+sec x =p { ((sec^2 x−tan^2 x=(p/4))),((tan^2 x+1−tan^2 x=(p/4)⇒p=4)) :} tan x+sec x = 4

$$\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2sin}\:\left(\frac{{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2sin}\:\left(\frac{{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$$\:\mathrm{sec}\:{x}−\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{4}}\:.\:{let}\:\mathrm{tan}\:{x}+\mathrm{sec}\:{x}\:={p} \\ $$$$\begin{cases}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{tan}\:^{\mathrm{2}} {x}=\frac{{p}}{\mathrm{4}}}\\{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}=\frac{{p}}{\mathrm{4}}\Rightarrow{p}=\mathrm{4}}\end{cases} \\ $$$$\:\:\mathrm{tan}\:{x}+\mathrm{sec}\:{x}\:=\:\mathrm{4} \\ $$

Commented by cortano last updated on 24/Dec/21

another way (((1−sin x)(1+sin x))/(cos x(1+sin x))) = (1/4) ((cos x)/(1+sin x)) = (1/4) ⇒((1+sin x)/(cos x))=4 ⇔ (1/(cos x))+tan x=4

$$\:{another}\:{way} \\ $$$$\:\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}{\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\mathrm{4} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+\mathrm{tan}\:{x}=\mathrm{4} \\ $$

Commented by peter frank last updated on 25/Dec/21

$$\mathrm{great} \\ $$

Commented by mnjuly1970 last updated on 24/Dec/21

$$\:\:{very}\:{nice}\:{solution}\:{sir}…{thank}\:{you} \\ $$

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