Question

x= 1+ cos theta/sin theta and y is= 1- cos theta /sin theta then find the relationship between x and y​

Answer 1

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

Given that

$\rm :\longmapsto\:x = \dfrac{1 + cos\theta }{sin\theta }$

can be rewritten as

$\rm :\longmapsto\:x =\dfrac{1}{sin\theta } + \dfrac{cos\theta }{sin\theta }$

We know,

$\green{ \boxed{ \sf{ \: \frac{1}{cosecx} = sinx}}} \: and \: \green{ \boxed{ \sf{ \: \frac{cosx}{sinx} = cotx}}}$

So, using this, we get

$\bf\implies \:x = cosec\theta + cot\theta - - - (1)$

Now, Consider

$\rm :\longmapsto\:y = \dfrac{1 - cos\theta }{sin\theta }$

$\rm :\longmapsto\: =\dfrac{1}{sin\theta } - \dfrac{cos\theta }{sin\theta }$

We know,

$\green{ \boxed{ \sf{ \: \frac{1}{cosecx} = sinx}}} \: and \: \green{ \boxed{ \sf{ \: \frac{cosx}{sinx} = cotx}}}$

So, using this,

$\bf\implies \:y = cosec\theta - cot\theta - - - (2)$

Now, we know that

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

We know,

$\red{ \boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

So, using this,

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

$\bf\implies \:xy = 1$

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