If sec x + tan x = 2+(√5) find sin x+ cos x ?


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Question Number 82654 by jagoll last updated on 23/Feb/20

$I f \sec x + \tan x = 2 + \sqrt{5}$ $find \sin x + \cos x ?$
Commented by jagoll last updated on 23/Feb/20

	Answered by mr W last updated on 23/Feb/20
	$((1+sin x)/(cos x))=2+(√5) cos x≠0 ⇒x≠2kπ±(π/2) (2+(√5))cos x−sin x=1 ((2+(√5))/(√(1^2 +(2+(√5))^2 ))) cos x−(1/(√(1^2 +(2+(√5))^2 ))) sin x=(1/(√(1^2 +(2+(√5))^2 ))) cos α cos x−sin α sin x=sin α cos (x+α)=sin α=cos ((π/2)−α) ⇒x+α=2kπ±((π/2)−α) ⇒x=2kπ±((π/2)−α)−α= { ((2kπ+(π/2)−2α)),((2kπ−(π/2) ⇒not suitable)) :} ⇒x=2kπ+(π/2)−2α cos x+sin x=sin 2α+cos 2α =((2(2+(√5)))/(10+4(√5)))+(((2+(√5))^2 −1)/(10+4(√5))) =((3(2+(√5)))/(5+2(√5))) =((3(2+(√5)))/((√5)(2+(√5)))) =((3(√5))/5)$
	$\frac{1 + \sin x}{\cos x} = 2 + \sqrt{5}$ $\cos x \neq 0 \Rightarrow x \neq 2 k π \pm \frac{π}{2}$ $(2 + \sqrt{5}) \cos x - \sin x = 1$ $\frac{2 + \sqrt{5}}{\sqrt{1^{2} + {(2 + \sqrt{5})}^{2}}} \cos x - \frac{1}{\sqrt{1^{2} + {(2 + \sqrt{5})}^{2}}} \sin x = \frac{1}{\sqrt{1^{2} + {(2 + \sqrt{5})}^{2}}}$ $\cos α \cos x - \sin α \sin x = \sin α$ $\cos (x + α) = \sin α = \cos (\frac{π}{2} - α)$ $\Rightarrow x + α = 2 k π \pm (\frac{π}{2} - α)$ $\Rightarrow x = 2 k π \pm (\frac{π}{2} - α) - α = {\begin{cases} 2 k π + \frac{π}{2} - 2 α \\ 2 k π - \frac{π}{2} \Rightarrow n o t s u i t a b l e \end{cases}$ $\Rightarrow x = 2 k π + \frac{π}{2} - 2 α$ $\cos x + \sin x = \sin 2 α + \cos 2 α$ $= \frac{2 (2 + \sqrt{5})}{10 + 4 \sqrt{5}} + \frac{{(2 + \sqrt{5})}^{2} - 1}{10 + 4 \sqrt{5}}$ $= \frac{3 (2 + \sqrt{5})}{5 + 2 \sqrt{5}}$ $= \frac{3 (2 + \sqrt{5})}{\sqrt{5} (2 + \sqrt{5})}$ $= \frac{3 \sqrt{5}}{5}$
	Commented by jagoll last updated on 23/Feb/20

	$t h a n k y o u s i r$
	Commented by mr W last updated on 23/Feb/20

	$i f p o s s i b l e o n e s h o u l d a l w a y s t r y$ $w i t h o u t s q u a r i n g t o a v o i d t h a t ‘ ‘ {f a k e}^{″}$ $r o o t s a r e a d d e d . w i t h x = 2 k π + \frac{π}{2} - 2 α$ $i^{'} m s u r e a l l t r u e r o o t s a r e i n c l u d e d a n d$ $n o f a k e r o o t i s i n c l u d e d .$
	Answered by TANMAY PANACEA last updated on 23/Feb/20

	$(s e c x + t a n x) (s e c x - t a n x) = 1$ $s e c x - t a n x = \frac{1}{\sqrt{5} + 2} = \sqrt{5} - 2$ $(s e c x + t a n x) + (s e c x - t a n x) = 2 \sqrt{5}$ $s e c x = \sqrt{5} \to c o s x = \frac{1}{\sqrt{5}}$ $s i n x = \frac{2}{\sqrt{5}} \to s i n x + c o s x = \frac{3}{\sqrt{5}} = \frac{3 \sqrt{5}}{5}$
	Commented by mr W last updated on 23/Feb/20

	$n i c e s o l u t i o n!$
	Commented by TANMAY PANACEA last updated on 23/Feb/20

	$t h a n k y o u s i r$
	Answered by john santu last updated on 23/Feb/20

	$l e t : \sec x - \tan x = t$ $(\sec x + \tan x) (\sec x - \tan x) = (2 + \sqrt{5}) t$ $\sec^{2} x - \tan^{2} x = (2 + \sqrt{5}) t$ $\tan^{2} x + 1 - \tan^{2} x = (2 + \sqrt{5}) t$ $t = \frac{1}{2 + \sqrt{5}} = \sqrt{5} - 2 = \frac{1}{\cos x} - \frac{\sin x}{\cos x} (1)$ $f r o m \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 2 + \sqrt{5} (2)$ $(1) + (2) \Rightarrow \frac{2}{\cos x} = 2 \sqrt{5} \Rightarrow \cos x = \frac{1}{\sqrt{5}}$ $t h e n \sin x = \frac{2}{\sqrt{5}} \Rightarrow \sin x + \cos x = \frac{3}{\sqrt{5}}$
	Commented by peter frank last updated on 23/Feb/20

	$t h a n k y o u b o t h$
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