Question

(Sec A+Tan A-1)(Sec A- Tan A+1)=2tanA

Answer 1

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:(secA + tanA - 1)(secA - tanA + 1) = 2tanA$

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

$\rm :\longmapsto\:(secA + tanA - 1)(secA - tanA + 1)$

$\rm \:  =  \:  \: \bigg(secA + (tanA - 1) \bigg) \bigg(secA - (tanA - 1) \bigg)$

We know,

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

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

We know,

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

So, using this, we get

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

We know,

$\boxed{ \bf{ \: 1 + {tan}^{2}x = {sec}^{2}x}}$

So, using this we get

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

$\rm \:  =  \:  \: {sec}^{2}A - {sec}^{2}A + 2tanA$

$\rm \:  =  \:  \: 2tanA$

Hence, Proved

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:

Given,

$\sf \: (secA + tanA - 1).(secA - tanA + 1) \\ \\ = \sf \{secA + (tanA - 1) \}. \{secA - (tanA - 1) \} \\ \\ \sf = {sec}^{2} A - {(tan}^{} A - 1) ^{2} \: \: \pink{\{ \because{x}^{2} - {y}^{2} = (x + y)(x - y) \} }\\ \\ \sf = {sec}^{2} A - \{ \red{{tan}^{2} A} - 2.tanA.1 + \red{ {1}^{2}} \} \\ \\ \sf = {sec}^{2} A - \{ \red{{sec}^{2} A} - 2tanA \} \: \: \: \: \red{\{\because \: {sec}^{2} A = 1 + {tan}^{2} A \}} \\ \\ \sf = \cancel{{sec}^{2} A }- \red{\cancel { {sec}^{2} A} }+ 2tanA \\ \\ \sf = 2tanA$

So,

L.H.S = R.H.S

Hence proved.