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Question Number 81963 by jagoll last updated on 17/Feb/20
if tan (x)+sec (x) = (7/8) find cot (x)+cosec (x) =
$${if}\:\mathrm{tan}\:\left({x}\right)+\mathrm{sec}\:\left({x}\right)\:=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${find}\:\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\:=\: \\ $$
Commented by john santu last updated on 17/Feb/20
let cot (x)+cosec (x) = t (i) {tan (x)+sec (x)}×{cot (x)+cosec (x)}= ((7t)/8) 1+ (1/(cos (x)))+(1/(sin (x)))+(1/(sin (x)cos (x))) = ((7t)/8) (ii) −((sin (x))/(cos (x)))−(1/(cos (x))) = −(7/8) (iii) −((cos (x))/(sin (x)))−(1/(sin (x))) = −t (i)+(ii)+(iii) ⇒ 1 = −(t/8)−(7/8) ⇒−8 = t+7 ⇒ t = −15 ∴ cot (x)+cosec (x) = −15
$${let}\:\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\:=\:{t} \\ $$$$\left({i}\right)\:\left\{\mathrm{tan}\:\left({x}\right)+\mathrm{sec}\:\left({x}\right)\right\}×\left\{\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\right\}=\:\frac{\mathrm{7}{t}}{\mathrm{8}} \\ $$$$\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}\right)\mathrm{cos}\:\left({x}\right)}\:=\:\frac{\mathrm{7}{t}}{\mathrm{8}} \\ $$$$\left({ii}\right)\:−\frac{\mathrm{sin}\:\left({x}\right)}{\mathrm{cos}\:\left({x}\right)}−\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}\right)}\:=\:−\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\left({iii}\right)\:−\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{sin}\:\left({x}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}\right)}\:=\:−{t} \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right) \\ $$$$\Rightarrow\:\mathrm{1}\:=\:−\frac{{t}}{\mathrm{8}}−\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow−\mathrm{8}\:=\:{t}+\mathrm{7}\:\Rightarrow\:{t}\:=\:−\mathrm{15}\: \\ $$$$\therefore\:\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\:=\:−\mathrm{15}\: \\ $$
Commented by jagoll last updated on 17/Feb/20
thank you mister
$${thank}\:{you}\:{mister} \\ $$
Answered by Kunal12588 last updated on 17/Feb/20
let tan x = t sec x = (√(1+t^2 )) cot x = (1/t) csc x = ((√(1+t^2 ))/t) t+(√(1+t^2 ))=(7/8) ⇒1+t^2 =((49)/(64))+t^2 −(7/4)t ⇒(7/4)t=((49)/(64))−1=((−15)/(64)) ⇒t=((−15)/(112)) (√(1+t^2 ))=(7/8)−t=(7/8)+((15)/(112))=((98+15)/(112))=((113)/(112)) (1/t)+((√(1+t^2 ))/t)=((1+(√(1+t^2 )))/t)=((1+((113)/(112)))/((−15)/(112)))=−((225)/(15))=−15 cot x + csc x = −15 I think there should be one more value when x is in IV quadrant
$${let}\:\:{tan}\:{x}\:=\:{t} \\ $$$${sec}\:{x}\:=\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${cot}\:{x}\:=\:\frac{\mathrm{1}}{{t}} \\ $$$${csc}\:{x}\:=\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}} \\ $$$${t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{1}+{t}^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{64}}+{t}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{4}}{t} \\ $$$$\Rightarrow\frac{\mathrm{7}}{\mathrm{4}}{t}=\frac{\mathrm{49}}{\mathrm{64}}−\mathrm{1}=\frac{−\mathrm{15}}{\mathrm{64}} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{15}}{\mathrm{112}} \\ $$$$\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{8}}−{t}=\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{15}}{\mathrm{112}}=\frac{\mathrm{98}+\mathrm{15}}{\mathrm{112}}=\frac{\mathrm{113}}{\mathrm{112}} \\ $$$$\frac{\mathrm{1}}{{t}}+\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}=\frac{\mathrm{1}+\frac{\mathrm{113}}{\mathrm{112}}}{\frac{−\mathrm{15}}{\mathrm{112}}}=−\frac{\mathrm{225}}{\mathrm{15}}=−\mathrm{15} \\ $$$${cot}\:{x}\:+\:{csc}\:{x}\:=\:−\mathrm{15} \\ $$$${I}\:{think}\:{there}\:{should}\:{be}\:{one}\:{more}\:{value}\:{when} \\ $$$${x}\:{is}\:{in}\:{IV}\:{quadrant} \\ $$

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