Question

$if \: \tan \theta = \frac{a}{b} \\ \\ \: find \: the \: value \: of \: \frac{ \cos \theta + \sin \theta}{ \cos \theta - \sin \theta }$
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Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{tan\theta \: = \dfrac{a}{b} }\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To\:find - \begin{cases} &\sf{\dfrac{cos\theta \: + sin\theta \:}{cos\theta \: - sin\theta \:} }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:Formula \: Used - \begin{cases} &\sf{tan\theta \: = \dfrac{sin\theta \:}{cos\theta \:} }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{cos\theta \: + sin\theta \:}{cos\theta \: - sin\theta \:}$

$\sf \: Divide \: numerator \: and \: denominator \: by \: cos\theta \:$

$\rm :\longmapsto\: = \: \dfrac{\dfrac{cos\theta \:}{cos\theta \:} + \dfrac{sin\theta \:}{cos\theta \:} }{\dfrac{cos\theta \:}{cos\theta \:} + \dfrac{sin\theta \:}{cos\theta \:} }$

$\rm :\longmapsto\: = \: \dfrac{1 + tan\theta \:}{1 - tan\theta \:}$

$\rm :\longmapsto\: = \: \dfrac{1 + \dfrac{a}{b} }{1 - \dfrac{a}{b} }$

$\rm :\longmapsto\: = \: \dfrac{b + a}{b - a}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1