Question

23 Using the identity sec? A = 1 + tan’A, prove that (sin A-cos A +1 )/(1 sin A+cos A-1) = 1/ (sec A-tan A)​

Answer 1

Appropriate Question :-

Using the identity,

$\sf \: {sec}^{2}A = 1 + {tan}^{2}A, \: prove \: that \: \dfrac{sinA - cosA + 1}{sinA + cosA - 1} = \dfrac{1}{secA - tanA}$

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

Consider LHS

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

Divide numerator and denominator by cosA, we get

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

$\red{ \boxed{ \sf{ \: \because \: tanx = \frac{sinx}{cosx} \: \: \: and \: \: \: \frac{1}{cosx} = secx}}}$

can be re-arranged as

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

We know,

$\red{ \boxed{ \sf{ \: {sec}^{2}x - {tan}^{2}x = 1}}}$

So, using this identity, we get

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

We know,

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

So, using this identity, we get

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

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

$\rm \:  =  \: \dfrac{(secA + tanA) \: \cancel{\bigg[1 - secA + tanA\bigg]}}{ \cancel{1 - secA + tanA}}$

$\rm \:  =  \: secA + tanA$

$\rm \:  =  \: secA + tanA \times \red{\dfrac{secA - tanA}{secA - tanA} }$

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

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

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