Question

tanA+secA-1/tanA-secA+1 = 1+sinA/cosA = cosA/1-sinA

Answer 1

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

Consider LHS

$\rm :\longmapsto\:\dfrac{tanA + secA - 1}{tanA - secA + 1}$

We know,

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

So, replace 1 of numerator by this, we get

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

We know, .

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

So, using this, we get

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

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

$\rm \:  =  \: secA + tanA$

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

$\red{\rm \:  =  \: \dfrac{1 + sinA}{cosA} }$

On rationalizing the numerator, we get

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

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{1 - {sin}^{2} A}{cosA(1 - sinA)}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: \dfrac{{cos}^{2} A}{cosA(1 - sinA)}$

$\red{\rm \:  =  \: \dfrac{cosA}{1 - sinA} }$

Hence, we concluded that

$\underbrace{\boxed{\tt{ \dfrac{tanA + secA - 1}{tanA - secA + 1} = \: \dfrac{1 + sinA}{cosA} \: = \: \dfrac{cosA}{1 - sinA}}}}$

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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

Answer:

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

Consider LHS

$\rm :\longmapsto\:\dfrac{tanA + secA - 1}{tanA - secA + 1}$

We know,

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

So, replace 1 of numerator by this, we get

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

We know, .

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

So, using this, we get

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

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

$\rm \:  =  \: secA + tanA$

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

$\red{\rm \:  =  \: \dfrac{1 + sinA}{cosA} }$

On rationalizing the numerator, we get

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

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{1 - {sin}^{2} A}{cosA(1 - sinA)}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: \dfrac{{cos}^{2} A}{cosA(1 - sinA)}$

$\red{\rm \:  =  \: \dfrac{cosA}{1 - sinA} }$

Hence, we concluded that

$\underbrace{\boxed{\tt{ \dfrac{tanA + secA - 1}{tanA - secA + 1} = \: \dfrac{1 + sinA}{cosA} \: = \: \dfrac{cosA}{1 - sinA}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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