sin-A-cos-A-1-sin-A-cos-A-1-

Question Number 112672 by bemath last updated on 09/Sep/20

$$\:\frac{\mathrm{sin}\:\mathrm{A}−\mathrm{cos}\:\mathrm{A}+\mathrm{1}}{\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}−\mathrm{1}}\:?\: \\ $$

Commented by som(math1967) last updated on 09/Sep/20

$$\frac{\mathrm{1}+\mathrm{sinA}}{\mathrm{cosA}} \\ $$

Commented by bemath last updated on 09/Sep/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all}.\:\mathrm{all}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct} \\ $$

Answered by bobhans last updated on 09/Sep/20

(⧫)(((sin A−cos A)+1)/((sin A+cos A)−1)) ×(((sin A+cos A)+1)/((sin A+cos A)+1)) = ((sin^2 A−cos^2 A+sin A−cos A+sin A+cos A+1)/(1+2sin Acos A−1)) = ((sin^2 A−(1−sin^2 A)+2sin A+1)/(2sin Acos A)) = ((2sin^2 A+2sin A)/(2sin Acos A)) = ((2sin A(sin A+1))/(2sin Acos A)) = ((sin A+1)/(cos A)) or ((sin A+1)/(cos A)) ×((sin A−1)/(sin A−1)) = ((sin^2 A−1)/(cos A(sin A−1))) = ((cos A)/(1−sin A))

$$\left(\blacklozenge\right)\frac{\left(\mathrm{sin}\:\mathrm{A}−\mathrm{cos}\:\mathrm{A}\right)+\mathrm{1}}{\left(\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}\right)−\mathrm{1}}\:×\frac{\left(\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}\right)+\mathrm{1}}{\left(\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}\right)+\mathrm{1}}\:= \\ $$$$\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}+\mathrm{sin}\:\mathrm{A}−\mathrm{cos}\:\mathrm{A}+\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}+\mathrm{1}}{\mathrm{1}+\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{A}−\mathrm{1}}\:= \\ $$$$\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}−\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}\right)+\mathrm{2sin}\:\mathrm{A}+\mathrm{1}}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{A}}\:= \\ $$$$\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{A}+\mathrm{2sin}\:\mathrm{A}}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{A}}\:=\:\frac{\mathrm{2sin}\:\mathrm{A}\left(\mathrm{sin}\:\mathrm{A}+\mathrm{1}\right)}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{A}} \\ $$$$=\:\frac{\mathrm{sin}\:\mathrm{A}+\mathrm{1}}{\mathrm{cos}\:\mathrm{A}}\:\mathrm{or}\:\frac{\mathrm{sin}\:\mathrm{A}+\mathrm{1}}{\mathrm{cos}\:\mathrm{A}}\:×\frac{\mathrm{sin}\:\mathrm{A}−\mathrm{1}}{\mathrm{sin}\:\mathrm{A}−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}−\mathrm{1}}{\mathrm{cos}\:\mathrm{A}\left(\mathrm{sin}\:\mathrm{A}−\mathrm{1}\right)}\:=\:\frac{\mathrm{cos}\:\mathrm{A}}{\mathrm{1}−\mathrm{sin}\:\mathrm{A}} \\ $$

Commented by som(math1967) last updated on 09/Sep/20

((cosA(1+sinA))/(cos^2 A))=((1+sinA)/(cosA))

$$\frac{\mathrm{cosA}\left(\mathrm{1}+\mathrm{sinA}\right)}{\mathrm{cos}^{\mathrm{2}} \mathrm{A}}=\frac{\mathrm{1}+\mathrm{sinA}}{\mathrm{cosA}} \\ $$

Commented by bemath last updated on 09/Sep/20

same sir. ((1+sin A)/(cos A)) = ((cos A)/(1−sin A))

$$\mathrm{same}\:\mathrm{sir}.\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{A}}{\mathrm{cos}\:\mathrm{A}}\:=\:\frac{\mathrm{cos}\:\mathrm{A}}{\mathrm{1}−\mathrm{sin}\:\mathrm{A}} \\ $$

Commented by som(math1967) last updated on 09/Sep/20

$$\mathrm{yes}\:\mathrm{sir} \\ $$

Answered by Dwaipayan Shikari last updated on 09/Sep/20

((2sin(A/2)cos(A/2)+2sin^2 (A/2))/(2sin(A/2)(cos(A/2)−sin(A/2))))=(((cos(A/2)+sin(A/2))^2 )/((cos^2 (A/2)−sin^2 (A/2))))=((1+sinA)/(cosA))=tanA+secA

$$\frac{\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}+\mathrm{2}{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}\left({cos}\frac{{A}}{\mathrm{2}}−{sin}\frac{{A}}{\mathrm{2}}\right)}=\frac{\left({cos}\frac{{A}}{\mathrm{2}}+{sin}\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left({cos}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}−{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\right)}=\frac{\mathrm{1}+{sinA}}{{cosA}}={tanA}+{secA} \\ $$

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