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Question Number 86486 by jagoll last updated on 29/Mar/20

If sin x + cos x = (2/3) find (1/(sin x)) + (1/(cos x)) = ?

$$\mathrm{If}\:\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{find}\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\:=\:? \\ $$

Commented by john santu last updated on 29/Mar/20

⇒ ((sin x+cos x)/(sin x cos x)) = (i) ⇒ (sin x+cos x)^2 = (4/9) 1+2sin x cos x = (4/9) sin x cos x = −(5/(18)) ⇒(i) = ((2/3)/(−(5/(18)))) =−(2/3)×((18)/5) = −((12)/5)

$$\Rightarrow\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}\:=\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{1}+\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:=\:−\frac{\mathrm{5}}{\mathrm{18}} \\ $$$$\Rightarrow\left(\mathrm{i}\right)\:=\:\frac{\frac{\mathrm{2}}{\mathrm{3}}}{−\frac{\mathrm{5}}{\mathrm{18}}}\:=−\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{18}}{\mathrm{5}} \\ $$$$=\:−\frac{\mathrm{12}}{\mathrm{5}} \\ $$

Commented by Serlea last updated on 29/Mar/20

(sinx+cosx)^2 =(4/9) sin^2 x+cos^2 x+sin2x=(4/9) 1+sin2x=(4/9) sin2x=((−5)/9) (1/(sinx))+(1/(cosx))=Y ((sinx+cosx)/(sinxcosx))=Y sinx+cosx=Ysinxcosx (2/3)=Ysinxcosx 2=3Ysinxcosx 2(2)=3y×2sinxcosx 4=3Ysin2x (4/(3Y))=((−5)/9) (4/y)=((−5)/3) y=−((12)/5)

$$\left(\mathrm{sinx}+\mathrm{cosx}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin2x}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{1}+\mathrm{sin2x}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{sin2x}=\frac{−\mathrm{5}}{\mathrm{9}} \\ $$$$\frac{\mathrm{1}}{\mathrm{sinx}}+\frac{\mathrm{1}}{\mathrm{cosx}}=\mathrm{Y} \\ $$$$\frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinxcosx}}=\mathrm{Y} \\ $$$$\mathrm{sinx}+\mathrm{cosx}=\mathrm{Ysinxcosx} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{Ysinxcosx} \\ $$$$\mathrm{2}=\mathrm{3Ysinxcosx} \\ $$$$\mathrm{2}\left(\mathrm{2}\right)=\mathrm{3y}×\mathrm{2sinxcosx} \\ $$$$\mathrm{4}=\mathrm{3Ysin2x} \\ $$$$\frac{\mathrm{4}}{\mathrm{3Y}}=\frac{−\mathrm{5}}{\mathrm{9}} \\ $$$$\frac{\mathrm{4}}{\mathrm{y}}=\frac{−\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{y}=−\frac{\mathrm{12}}{\mathrm{5}} \\ $$

Commented by jagoll last updated on 29/Mar/20

sin x + cos x = (2/3) sir not = 1

$$\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{sir} \\ $$$$\mathrm{not}\:=\:\mathrm{1} \\ $$

Commented by jagoll last updated on 29/Mar/20

something wrong

$$\mathrm{something}\:\mathrm{wrong} \\ $$

Commented by Serlea last updated on 29/Mar/20

Thanks I was correcting it already It was oversight

$$\mathrm{Thanks} \\ $$$$\mathrm{I}\:\mathrm{was}\:\mathrm{correcting}\:\mathrm{it}\:\mathrm{already} \\ $$$$\mathrm{It}\:\mathrm{was}\:\mathrm{oversight} \\ $$

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