Question

＠ChallengingModerators

Prove That :-
$\rm\dfrac{1 + cosA}{1 - cosA} = \dfrac{tan^{2}A}{(sec A - 1)^{2}}$

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Answer 1

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

Consider RHS

$\rm :\longmapsto\:\dfrac{tan^{2}A}{(sec A - 1)^{2}}$

We know,

$\boxed{\tt{ {sec}^{2}x - {tan}^{2}x = 1}}$

So, using this

$\rm \:  =  \: \dfrac{sec^{2}A - 1}{(sec A - 1)^{2}}$

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{(secA - 1)(secA + 1)}{(sec A - 1)^{2}}$

$\rm \:  =  \: \dfrac{secA + 1}{secA - 1}$

$\rm \:  =  \: \dfrac{\dfrac{1}{cosA} + 1}{\dfrac{1}{cosA} - 1}$

$\rm \:  =  \: \dfrac{\dfrac{1 + cosA}{cosA}}{\dfrac{1 - cosA}{cosA}}$

$\rm \:  =  \: \dfrac{1 + cosA}{1 - cosA}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{tan^{2}A}{(sec A - 1)^{2}} = \frac{1 + cosA}{1 - cosA}}}$

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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