Question

sin²Q(1/1-cosQ+1/1+cosQ)=2​

Answer 1

Step-by-step explanation:

Hope, will you understand.

Answer 2

Given Question

Prove that,

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

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

Consider, LHS

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

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

We know,

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

So, using this, we get

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

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

We know,

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

So, using this, we get

$\rm \:  =  \: {sin}^{2}\theta \bigg( \dfrac{2}{ {sin}^{2}\theta } \bigg)$

$\rm \:  =  \: 2$

Hence,

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

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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