sec x − cosec x=(√(35)) tan x+cot x=?


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

$\sec x - cosec x = \sqrt{35}$ $\tan x + \cot x = ?$
	Answered by john santu last updated on 11/Aug/20
	$⋇JS⋇ (1) (1/(cos x))−(1/(sin x)) = (√(35)) ⇒sin x−cos x = (√(35)) cos x sin x (2) let tan x+cot x = k ⇒((sin x)/(cos x))+((cos x)/(sin x)) = k cos x sin x = (1/k) (3) (sin x−cos x)^2 = ((35)/k^2 ) ⇒1−(2/k) = ((35)/k^2 ) ⇒((35)/k^2 )+(2/k)−1=0 ⇒−k^2 +2k+35 = 0 k^2 −2k−35 = 0 → { ((k=7)),((k=−5)) :} ∴ tan x+cot x = 7 or −5$
	$⋇ JS ⋇$ $(1) \frac{1}{\cos x} - \frac{1}{\sin x} = \sqrt{35}$ $\Rightarrow \sin x - \cos x = \sqrt{35} \cos x \sin x$ $(2) l e t \tan x + \cot x = k$ $\Rightarrow \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = k$ $\cos x \sin x = \frac{1}{k}$ $(3) {(\sin x - \cos x)}^{2} = \frac{35}{k^{2}}$ $\Rightarrow 1 - \frac{2}{k} = \frac{35}{k^{2}} \Rightarrow \frac{35}{k^{2}} + \frac{2}{k} - 1 = 0$ $\Rightarrow - k^{2} + 2 k + 35 = 0$ $k^{2} - 2 k - 35 = 0 \to {\begin{cases} k = 7 \\ k = - 5 \end{cases}$ $∴ \tan x + \cot x = 7 o r - 5$
	Commented by ZiYangLee last updated on 11/Aug/20

	$Wow Nice Solution!$
	Answered by $@y@m last updated on 11/Aug/20

	$\frac{1}{\cos x} - \frac{1}{\sin x} = \sqrt{35}$ $\sin x - \cos x = \sqrt{35} \sin x \cos x$ ${(\sin x - \cos x)}^{2} = 35 \sin^{2} x \cos^{2} x$ $1 - 2 \sin x \cos x = 35 \sin^{2} x \cos^{2} x$ $L e t \sin x \cos x = t$ $35 t^{2} + 2 t - 1 = 0$ $t = \frac{- 2 \pm \sqrt{4 + 140}}{70} = \frac{- 2 \pm 12}{70}$ $t = \frac{1}{7}, \frac{- 1}{5}$ $∴ \tan x + \cot x = 7, - 5$
	Answered by 1549442205PVT last updated on 11/Aug/20
	$(1/(cosx))−(1/(sinx))=(√(35))⇔(1/(cos^2 x))−(2/(cosxsinx))+(1/(sin^2 x))=35 ⇔((sin^2 x+cos^2 x)/(cos^2 xsin^2 x))−(2/(cosxsinx))=35 ⇔(1/(cos^2 xsin^2 x))−(2/(cosxsinx))=35.Set (1/(cosxsinx))=y we get y^2 −2y−35=0.⇔(y−7)(y+5)=0 ⇔y∈{−5;7}.On the other hands,we have tanx+cotx=((sinx)/(cosx))+((cosx)/(sinx)) ((sin^2 x+cos^2 x)/(cosxsinx))=(1/(cosxsinx)) Thus,tanx+cotx=y∈{−5;7}$
	$\frac{1}{cosx} - \frac{1}{sinx} = \sqrt{35} \Leftrightarrow \frac{1}{\cos^{2} x} - \frac{2}{cosxsinx} + \frac{1}{\sin^{2} x} = 35$ $\Leftrightarrow \frac{\sin^{2} x + \cos^{2} x}{\cos^{2} {xsin}^{2} x} - \frac{2}{cosxsinx} = 35$ $\Leftrightarrow \frac{1}{\cos^{2} {xsin}^{2} x} - \frac{2}{cosxsinx} = 35 . Set \frac{1}{cosxsinx} = y$ $we get y^{2} - 2 y - 35 = 0 . \Leftrightarrow (y - 7) (y + 5) = 0$ $\Leftrightarrow y \in {- 5; 7} . On the other hands, we have$ $tanx + cotx = \frac{sinx}{cosx} + \frac{cosx}{sinx}$ $\frac{\sin^{2} x + \cos^{2} x}{cosxsinx} = \frac{1}{cosxsinx}$ $Thus, tanx + cotx = y \in {- 5; 7}$
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