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$\tt{if} \: \frac{2sin \theta}{1 + cos \theta + sin \theta} = y. \:then \: \frac{1 - cos \theta + sin \theta}{1 + sin \theta} is \: also \: y.$
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Answer 1

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

Given that,

$\rm :\longmapsto\:y = \dfrac{2sin\theta }{1 + cos\theta + sin\theta }$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{2sin\theta }{1 + sin\theta + cos\theta }$

$\rm :\longmapsto\:y = \dfrac{2sin\theta }{(1 + sin\theta) + cos\theta }$

So, on rationalizing the denominator, we get

$\rm :\longmapsto\:y = \dfrac{2sin\theta }{(1 + sin\theta) + cos\theta } \times \dfrac{1 + sin\theta - cos\theta }{(1 + sin\theta ) - cos\theta }$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{(1 + sin\theta)^{2} - cos ^{2} \theta }$

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{1 + {sin}^{2}\theta+ 2sin\theta-{cos}^{2}\theta }$

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{(1 - {cos}^{2}\theta )+{sin}^{2}\theta+ 2sin\theta }$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{{sin}^{2}\theta +{sin}^{2}\theta+ 2sin\theta }$

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{2{sin}^{2}\theta+ 2sin\theta }$

$\rm :\longmapsto\:y = \dfrac{2sin\theta (1 + sin\theta - cos\theta )}{2sin\theta (sin\theta + 1) }$

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

Hence,

$\rm \implies\:\:\boxed{ \tt{ \: y = \dfrac{1 - cos\theta + sin\theta }{1 + sin\theta } \: }}$

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