Question

1+cos theta–sin^2theta/sintheta(1+costheta)=cottheta
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Answer 1

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

Consider,

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

can be rewritten as

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

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So using this, we get

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

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

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

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

$\rm \:  =  \: cot\theta$

Hence,

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

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

Answer 2

Answer:

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

Consider,

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

can be rewritten as

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

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So using this, we get

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

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

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

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

$\rm \:  =  \: cot\theta$

Hence,

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

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