Question

prove the idevtity 1+cos theta \1-cos theta = (cosec theta + cot theta)​

Answer 1

$\underline{ \underline{ \bf \: Given \: to \: prove \: that}}$

$\sqrt{\bf \dfrac{1 + cos \theta}{1 - cos \theta} = cosec \theta + cot \theta}$

Required Explanation :-

Take L.H.S

$\sqrt{\rm \dfrac{1 + cos \theta}{1 - cos \theta} }$

Multiplying and dividing with conjugate of denominator

Conjugate of 1-cosθ is 1+cosθ

$\sqrt{\rm \dfrac{1 + cos \theta}{1 - cos \theta} \times \rm \dfrac{1 + cos \theta}{1 + cos \theta} }$

$\sqrt{\rm \dfrac{(1 + cos \theta)(1 + cos \theta)}{(1 + cos \theta)( 1 - cos \theta)} }$

$\sqrt{\rm \dfrac{(1 + cos \theta) {}^{2} }{(1 + cos \theta)( 1 - cos \theta)} }$

$\bigg\{\rm\red{Numerator\: is \:in \: form \: (a + b)^2 = a^2+2ab+b^2}\bigg\}$

$\bigg\{\rm\red{Denominator\: is \:in \: form \: (a + b)(a-b) = a^2-b^2}\bigg\}$

$\sqrt{ \rm \dfrac{(1 ) {}^{2} + cos {}^{2} \theta + 2cos \theta }{1 - cos {}^{2} \theta} }$

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

Dividing with sin² theta for each term .

$\sqrt{\rm \: \dfrac{1}{sin {}^{2} \theta} + \dfrac{cos {}^{2} \theta }{sin {}^{2} \theta } + \dfrac{2cos \theta}{sin {}^{2} \theta} }$

$\sqrt{\rm \: cosec {}^{2} \theta + cot {}^{2} \theta + \dfrac{2cos \theta}{sin {} \theta} \times \dfrac{1}{sin \theta} }$

$\bigg \{ \red{ \rm\dfrac{1}{sin \theta} = csc \theta \: , \dfrac{cos \theta}{sin \theta} = cot \theta \: } \bigg \}$

$\sqrt{\rm \: cosec {}^{2} \theta + cot {}^{2} \theta + 2cot \theta \times cosec \theta }$

$\sqrt{\rm \: cosec {}^{} \theta + cot \theta) {}^{2} }$ $\bigg\{\rm\red{a^2+b^2+2ab= (a+b)^2}\bigg\}$

$= \rm \: csc \theta + cot \theta$

$\mathfrak\red{Hence \:proved !}$

Answer 2

Appropriate Question :

Prove the identity

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

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

Consider, LHS

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

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1 + cos\theta }{1 - cos\theta } \times \dfrac{1 + cos\theta }{1 + cos\theta }$

We know

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

So, using this, we get

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

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

We know

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

So, using this identity, we get

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

$\rm \: = \: {\bigg[\dfrac{1 + cos\theta }{sin\theta } \bigg]}^{2}$

$\rm \: = \: {\bigg[ \dfrac{1}{sin\theta } + \dfrac{cos\theta }{sin\theta } \bigg]}^{2}$

$\rm \: = \: {(cosec\theta + cot\theta )}^{2}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{1 + cos\theta }{1 - cos\theta } = {(cosec\theta + cot\theta )}^{2} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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