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Question Number 103643 by bemath last updated on 16/Jul/20

if sin x+cos x = (5/6) then (1/(sin x)) + (1/(cos x)) ?

$${if}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${then}\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:?\: \\ $$

Answered by Dwaipayan Shikari last updated on 16/Jul/20

(sinx+cosx)^2 −2sinxcosx=1 ((25)/(36))−2sinxcosx=1 sinxcosx=−((11)/(72)) (1/(sinx))+(1/(cosx))=((5/6)/((−11)/(72)))=−((60)/(11))

$$\left({sinx}+{cosx}\right)^{\mathrm{2}} −\mathrm{2}{sinxcosx}=\mathrm{1} \\ $$$$\frac{\mathrm{25}}{\mathrm{36}}−\mathrm{2}{sinxcosx}=\mathrm{1} \\ $$$${sinxcosx}=−\frac{\mathrm{11}}{\mathrm{72}} \\ $$$$\frac{\mathrm{1}}{{sinx}}+\frac{\mathrm{1}}{{cosx}}=\frac{\frac{\mathrm{5}}{\mathrm{6}}}{\frac{−\mathrm{11}}{\mathrm{72}}}=−\frac{\mathrm{60}}{\mathrm{11}} \\ $$

Answered by bobhans last updated on 16/Jul/20

(1/(sin x))+(1/(cos x)) = p ...(2) (1)×(2) ⇒ (sin x+cos x) ((1/(sin x))+(1/(cos x)))=((5p)/6) ⇒1+((sin x)/(cos x))+((cos x)/(sin x))+1 = ((5p)/6) (1/(cos x sin x))+2 = ((5p)/6) ...(3) (sin x+cos x)^2 = ((25)/(36)) 2sin xcos x = −((11)/(36))⇒sin x cos x = −((11)/(72)) now eq (3) ⇒ −((72)/(11))+2 = ((5p)/6) ((5p)/6) = −((50)/(11)) ⇒p = −((60)/(11)) ★

$$\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\:{p}\:...\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)\:\Rightarrow\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right)=\frac{\mathrm{5}{p}}{\mathrm{6}}\: \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}+\mathrm{1}\:=\:\frac{\mathrm{5}{p}}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}+\mathrm{2}\:=\:\frac{\mathrm{5}{p}}{\mathrm{6}}\:...\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{25}}{\mathrm{36}} \\ $$$$\mathrm{2sin}\:{x}\mathrm{cos}\:{x}\:=\:−\frac{\mathrm{11}}{\mathrm{36}}\Rightarrow\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:=\:−\frac{\mathrm{11}}{\mathrm{72}} \\ $$$${now}\:{eq}\:\left(\mathrm{3}\right)\:\Rightarrow\:−\frac{\mathrm{72}}{\mathrm{11}}+\mathrm{2}\:=\:\frac{\mathrm{5}{p}}{\mathrm{6}} \\ $$$$\frac{\mathrm{5}{p}}{\mathrm{6}}\:=\:−\frac{\mathrm{50}}{\mathrm{11}}\:\Rightarrow{p}\:=\:−\frac{\mathrm{60}}{\mathrm{11}}\:\bigstar \\ $$

Answered by mathmax by abdo last updated on 16/Jul/20

sinx +cosx =(5/6) ⇒(sinx +cosx)^2 =((25)/(36)) ⇒1 +2sinx cosx =((25)/(36)) ⇒ 2sinx cosx =((25)/(36))−1 =−((11)/(36)) ⇒sinx cosx =−((11)/(72)) ⇒ (1/(sinx))+(1/(cosx)) =((cosx +sinx)/(sinx cosx)) =((5/6)/(−((11)/(72)))) =−(5/6)×((72)/(11)) =−((5.24.3)/(2.3.11)) =−((5.12)/(11)) =−((60)/(11))

$$\mathrm{sinx}\:+\mathrm{cosx}\:=\frac{\mathrm{5}}{\mathrm{6}}\:\Rightarrow\left(\mathrm{sinx}\:+\mathrm{cosx}\right)^{\mathrm{2}} \:=\frac{\mathrm{25}}{\mathrm{36}}\:\Rightarrow\mathrm{1}\:+\mathrm{2sinx}\:\mathrm{cosx}\:=\frac{\mathrm{25}}{\mathrm{36}}\:\Rightarrow \\ $$$$\mathrm{2sinx}\:\mathrm{cosx}\:=\frac{\mathrm{25}}{\mathrm{36}}−\mathrm{1}\:=−\frac{\mathrm{11}}{\mathrm{36}}\:\Rightarrow\mathrm{sinx}\:\mathrm{cosx}\:=−\frac{\mathrm{11}}{\mathrm{72}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{sinx}}+\frac{\mathrm{1}}{\mathrm{cosx}}\:=\frac{\mathrm{cosx}\:+\mathrm{sinx}}{\mathrm{sinx}\:\mathrm{cosx}}\:=\frac{\frac{\mathrm{5}}{\mathrm{6}}}{−\frac{\mathrm{11}}{\mathrm{72}}}\:=−\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{72}}{\mathrm{11}}\:=−\frac{\mathrm{5}.\mathrm{24}.\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{11}}\:=−\frac{\mathrm{5}.\mathrm{12}}{\mathrm{11}}\:=−\frac{\mathrm{60}}{\mathrm{11}} \\ $$$$ \\ $$

Answered by Aziztisffola last updated on 16/Jul/20

(1/(sin x)) + (1/(cos x)) =((5/6)/(sin x.cos x)) (cosx+sinx)^2 =1+2sinxcosx=((25)/(36)) sinxcosx=((((25)/(36))−1)/2)=((−((11)/(36)))/2)=−((11)/(72)) (1/(sin x)) + (1/(cos x)) =((5/6)/(−((11)/(72))))=(5/6)×((72)/(−11))=−((360)/(66))

$$\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:=\frac{\frac{\mathrm{5}}{\mathrm{6}}}{\mathrm{sin}\:\mathrm{x}.\mathrm{cos}\:\mathrm{x}} \\ $$$$\:\left(\mathrm{cosx}+\mathrm{sinx}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2sinxcosx}=\frac{\mathrm{25}}{\mathrm{36}} \\ $$$$\:\mathrm{sinxcosx}=\frac{\frac{\mathrm{25}}{\mathrm{36}}−\mathrm{1}}{\mathrm{2}}=\frac{−\frac{\mathrm{11}}{\mathrm{36}}}{\mathrm{2}}=−\frac{\mathrm{11}}{\mathrm{72}} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:=\frac{\frac{\mathrm{5}}{\mathrm{6}}}{−\frac{\mathrm{11}}{\mathrm{72}}}=\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{72}}{−\mathrm{11}}=−\frac{\mathrm{360}}{\mathrm{66}} \\ $$$$\: \\ $$

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