Question

(cos theta+sin theta) :(cos theta-sin theta) =(√3+1): (√3-1)
Then find sec theta

please answer seriously...​

Answer 1

Answer:

$\bold{sec \alpha = \dfrac{2}{ \sqrt{3} }}$

Step-by-step explanation:

$\bold{Given} = > \\ \dfrac{cos \alpha + sin \alpha }{cos \alpha - sin \alpha } = \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1 }$

$to \: find = > value \: of \: sec \alpha$

$\bold{Concept \: used} = > if \\ \dfrac{a}{b} = \dfrac{c}{d} \\ applying \: componendo \: and \: dividendo \\ = > \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

$\bold{Solution} = > \\ \dfrac{cos \alpha + sin \alpha }{cos \alpha - sin \alpha } = \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1}$

$applying \: componendo \: and \: dividendo \:$

$= > \dfrac{cos \alpha + sin \alpha + cos \alpha - sin \alpha }{cos \alpha + sin \alpha - cos \alpha + sin \alpha } = \dfrac{ \sqrt{3 } + 1 + \sqrt{3} - 1 }{ \sqrt{3} + 1 - \sqrt{3} + 1 }$

$= > \dfrac{2cos \alpha }{2sin \alpha } = \dfrac{2 \sqrt{3} }{2}$

$= > \dfrac{cos \alpha }{sin \alpha } = \dfrac{ \sqrt{3} }{1}$

$= > cot \alpha = \sqrt{3}$

$= > cot \alpha = cot \: 30^{0}$

$= > \: \: \alpha = 30 ^{0}$

$= > sec \alpha =s ec30 ^{0}$

$= > sec \alpha = \dfrac{2}{ \sqrt{3} }$

Answer 2

see to the attachment given ......