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Question Number 107471 by ZiYangLee last updated on 11/Aug/20

sec x − cosec x=(√(35)) tan x+cot x=?

$$\mathrm{sec}\:\mathrm{x}\:−\:\mathrm{cosec}\:\mathrm{x}=\sqrt{\mathrm{35}} \\ $$$$\mathrm{tan}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}=? \\ $$

Answered by john santu last updated on 11/Aug/20

⋇JS⋇ (1) (1/(cos x))−(1/(sin x)) = (√(35)) ⇒sin x−cos x = (√(35)) cos x sin x (2) let tan x+cot x = k ⇒((sin x)/(cos x))+((cos x)/(sin x)) = k cos x sin x = (1/k) (3) (sin x−cos x)^2 = ((35)/k^2 ) ⇒1−(2/k) = ((35)/k^2 ) ⇒((35)/k^2 )+(2/k)−1=0 ⇒−k^2 +2k+35 = 0 k^2 −2k−35 = 0 → { ((k=7)),((k=−5)) :} ∴ tan x+cot x = 7 or −5

$$\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:=\:\sqrt{\mathrm{35}}\: \\ $$$$\Rightarrow\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:=\:\sqrt{\mathrm{35}}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:{x} \\ $$$$\left(\mathrm{2}\right)\:{let}\:\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\:=\:{k}\: \\ $$$$\Rightarrow\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:=\:{k}\: \\ $$$$\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{1}}{{k}}\: \\ $$$$\left(\mathrm{3}\right)\:\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\:\frac{\mathrm{35}}{{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}}{{k}}\:=\:\frac{\mathrm{35}}{{k}^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{35}}{{k}^{\mathrm{2}} }+\frac{\mathrm{2}}{{k}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow−{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{35}\:=\:\mathrm{0}\: \\ $$$${k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{35}\:=\:\mathrm{0}\:\rightarrow\begin{cases}{{k}=\mathrm{7}}\\{{k}=−\mathrm{5}}\end{cases} \\ $$$$\therefore\:\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\:=\:\mathrm{7}\:{or}\:−\mathrm{5}\: \\ $$

Commented by ZiYangLee last updated on 11/Aug/20

Wow Nice Solution!

$$\mathrm{Wow}\:\mathrm{Nice}\:\mathrm{Solution}! \\ $$

Answered by $@y@m last updated on 11/Aug/20

(1/(cos x))−(1/(sin x))=(√(35)) sin x−cos x=(√(35))sin xcos x (sin x−cos x)^2 =35sin^2 xcos^2 x 1−2sin xcos x=35sin^2 xcos^2 x Let sin xcos x=t 35t^2 +2t−1=0 t=((−2±(√(4+140)))/(70)) =((−2±12)/(70)) t=(1/7),((−1)/5) ∴ tan x+cot x=7, −5

$$\frac{\mathrm{1}}{\mathrm{cos}\:\:{x}}−\frac{\mathrm{1}}{\mathrm{sin}\:\:{x}}=\sqrt{\mathrm{35}} \\ $$$$\mathrm{sin}\:{x}−\mathrm{cos}\:{x}=\sqrt{\mathrm{35}}\mathrm{sin}\:{x}\mathrm{cos}\:{x} \\ $$$$\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\mathrm{35sin}\:^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{1}−\mathrm{2sin}\:{x}\mathrm{cos}\:{x}=\mathrm{35sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$${Let}\:\mathrm{sin}\:{x}\mathrm{cos}\:{x}={t} \\ $$$$\mathrm{35}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{140}}}{\mathrm{70}}\:=\frac{−\mathrm{2}\pm\mathrm{12}}{\mathrm{70}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{7}},\frac{−\mathrm{1}}{\mathrm{5}} \\ $$$$\therefore\:\mathrm{tan}\:{x}+\mathrm{cot}\:{x}=\mathrm{7},\:−\mathrm{5} \\ $$

Answered by 1549442205PVT last updated on 11/Aug/20

(1/(cosx))−(1/(sinx))=(√(35))⇔(1/(cos^2 x))−(2/(cosxsinx))+(1/(sin^2 x))=35 ⇔((sin^2 x+cos^2 x)/(cos^2 xsin^2 x))−(2/(cosxsinx))=35 ⇔(1/(cos^2 xsin^2 x))−(2/(cosxsinx))=35.Set (1/(cosxsinx))=y we get y^2 −2y−35=0.⇔(y−7)(y+5)=0 ⇔y∈{−5;7}.On the other hands,we have tanx+cotx=((sinx)/(cosx))+((cosx)/(sinx)) ((sin^2 x+cos^2 x)/(cosxsinx))=(1/(cosxsinx)) Thus,tanx+cotx=y∈{−5;7}

$$\frac{\mathrm{1}}{\mathrm{cosx}}−\frac{\mathrm{1}}{\mathrm{sinx}}=\sqrt{\mathrm{35}}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{2}}{\mathrm{cosxsinx}}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\mathrm{35} \\ $$$$\Leftrightarrow\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{2}}{\mathrm{cosxsinx}}=\mathrm{35} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{2}}{\mathrm{cosxsinx}}=\mathrm{35}.\mathrm{Set}\:\frac{\mathrm{1}}{\mathrm{cosxsinx}}=\mathrm{y} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{y}^{\mathrm{2}} −\mathrm{2y}−\mathrm{35}=\mathrm{0}.\Leftrightarrow\left(\mathrm{y}−\mathrm{7}\right)\left(\mathrm{y}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{y}\in\left\{−\mathrm{5};\mathrm{7}\right\}.\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands},\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{tanx}+\mathrm{cotx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}+\frac{\mathrm{cosx}}{\mathrm{sinx}} \\ $$$$\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{cosxsinx}}=\frac{\mathrm{1}}{\mathrm{cosxsinx}} \\ $$$$\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{tanx}}+\boldsymbol{\mathrm{cotx}}=\boldsymbol{\mathrm{y}}\in\left\{−\mathrm{5};\mathrm{7}\right\} \\ $$

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