Question

$\sqrt { \dfrac { 1 - \cos \theta } { 1 + \cos \theta } } =$​

Answer 1

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

Consider,

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

So, on rationalizing the denominator, we get

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

We know

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

So, using this identity, we get

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

$\rm \: = \: \dfrac{1 - cos\theta }{ \sqrt{1 - {cos}^{2}\theta } }$

We know,

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

So, using this identity, we get

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

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

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

$\rm \: = \: cosec\theta \: - \: cot\theta$

Hence,

$\rm\implies \: \boxed{\tt{ \: \sqrt{ \frac{1 - cos\theta }{1 + cos\theta } } = \: cosec\theta \: - \: cot\theta \: }}$

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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

$\bold{ \: solution }\\ \to consider \\ \ \sqrt{ \frac{1 - cos \theta}{1 + cos \theta} } \\ \\ \sf \: so \: on \: rationalizing \: the \: denominator \\ \sf \: we \: get \\ \\ : \to \sqrt{ \frac{1 - cos \theta}{1 + cos \theta} \times \: \frac{1 - \cos \theta }{1 - \cos \theta } } \\ \\ \bold{ : : we \: \: know} \\ \\ \boxed{(x - y)(x + y) = {x}^{2} - {y}^{2} } \\ \\ \sf \: so \: using \: the \: identity \: we \: get \\ \to \: \frac{1 - \cos \theta }{ \sqrt{ {sin}^{2} \theta} } \\ \\ \sf \: so \: using \: this \: identity \: we \: get \\ \\ \to \: \sqrt{ \frac{(1 - \cos \theta) ^{2} }{ {1}^{2} - cos ^{2} \theta} } \\ \\ \to \: \frac{1 - cos \theta}{ \sqrt{1 - {cos}^{2} \theta} } \\ \\ \bold{we \: know} \\ \\ \to \: sin ^{2} \theta \: + {cos}^{2} \theta \: = 1 \\ \\ \sf \: so \: using \: this \: identity \: we \: get \:$

= 1/-cos∅/√sin²∅

= 1-cos∅/sin∅

= 1/sin∅ - cos∅/sin∅

= cosec∅ - cot∅

Hence

=

$: \to \sqrt{ \frac{1 - cos \theta}{1 + cos \theta} } = cosec\theta \: - cot \theta$