If-sec-x-tan-x-2-5-find-sin-x-cos-x-

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

\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