Question

CSC theta -cot theta =1/3 ,then csc theta +cot theta ?

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{cosec\theta \: - cot\theta \: = \dfrac{1}{3} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To\:find - \begin{cases} &\sf{cosec\theta \: + cot\theta \:}  \end{cases}\end{gathered}\end{gathered}$

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

Given that,

$\rm :\longmapsto\:cosec\theta \: - cot\theta \: = \dfrac{1}{3} - - - (1)$

We know that,

$\rm :\longmapsto\: {cosec}^{2} \theta \: - {cot}^{2} \theta \: = 1$

$\rm :\longmapsto\:(cosec\theta \: - cot\theta \:)(cosec\theta \: + cot\theta \:) = 1$

$\rm :\longmapsto\:\dfrac{1}{3} \times (cosec\theta \: + cot\theta \:) = 1 \: \: \: \: \: \: \: \: \{ \because \: using \: (1) \}$

$\rm :\longmapsto\:cosec\theta \: + cot\theta \: = 3$

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