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Question Number 161903 by amin96 last updated on 23/Dec/21

((1−sin(x)−cos(x))/(1+sin(x)−cos(x)))=???

$$\frac{\mathrm{1}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}=??? \\ $$$$ \\ $$

Commented by blackmamba last updated on 24/Dec/21

=((1−2sin (1/2)x cos (1/2)x−(1−2sin^2 (1/2)x))/(1+2sin (1/2)x cos (1/2)x−(1−2sin^2 (1/2)x))) =((2sin (1/2)x (sin (1/2)x−cos (1/2)x))/(2sin (1/2)x (sin (1/2)x+cos (1/2)x))) = ((sin (1/2)x−cos (1/2)x)/(sin (1/2)x+cos (1/2)x)) = ((tan (1/2)x−1)/(tan (1/2)x+1)) = tan ((1/2)x−(π/4))

$$\:=\frac{\mathrm{1}−\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{1}+\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\ $$$$\:=\frac{\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\left(\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\left(\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\ $$$$\:=\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:=\:\frac{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{1}}{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}} \\ $$$$\:=\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$

Answered by MJS_new last updated on 23/Dec/21

tan x −sec x !!!!!

$$\mathrm{tan}\:{x}\:−\mathrm{sec}\:{x}\:!!!!! \\ $$

Answered by Ar Brandon last updated on 24/Dec/21

f(x)=((1−sinx−cosx)/(1+sinx−cosx))=((2sin^2 (x/2)−2sin(x/2)cos(x/2))/(2sin^2 (x/2)+2sin(x/2)cos(x/2))) =((2sin(x/2)(sin(x/2)−cos(x/2)))/(2sin(x/2)(sin(x/2)+cos(x/2))))=((sin(x/2)−cos(x/2))/(sin(x/2)+cos(x/2))) =(((sin(x/2)−cos(x/2))^2 )/(sin^2 (x/2)−cos^2 (x/2)))=((sin^2 (x/2)−2sin(x/2)cos(x/2)+cos^2 (x/2))/(−cosx)) =((sinx−1)/(cosx))=tanx−secx

$${f}\left({x}\right)=\frac{\mathrm{1}−\mathrm{sin}{x}−\mathrm{cos}{x}}{\mathrm{1}+\mathrm{sin}{x}−\mathrm{cos}{x}}=\frac{\mathrm{2sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2sin}\frac{{x}}{\mathrm{2}}\left(\mathrm{sin}\frac{{x}}{\mathrm{2}}−\mathrm{cos}\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2sin}\frac{{x}}{\mathrm{2}}\left(\mathrm{sin}\frac{{x}}{\mathrm{2}}+\mathrm{cos}\frac{{x}}{\mathrm{2}}\right)}=\frac{\mathrm{sin}\frac{{x}}{\mathrm{2}}−\mathrm{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{sin}\frac{{x}}{\mathrm{2}}+\mathrm{cos}\frac{{x}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\frac{{x}}{\mathrm{2}}−\mathrm{cos}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}=\frac{\mathrm{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}+\mathrm{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{−\mathrm{cos}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}{x}−\mathrm{1}}{\mathrm{cos}{x}}=\mathrm{tan}{x}−\mathrm{sec}{x} \\ $$

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

great

$$\mathrm{great} \\ $$

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