Question

Prove that:

sec A + tan A /sec A - tan A =[1 + sin A /cos A]²​

Answer 1

Given Question :-

$\sf \: Prove \: that: \dfrac{secA + tanA}{secA - tanA} = {\bigg(\dfrac{1 + sinA}{cosA} \bigg) }^{2}$

Identities Used :-

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

$\purple{\boxed{ \bf \:secx = \dfrac{1}{cosx}}}$

$\purple{\boxed{ \bf \:tanx = \dfrac{sinx}{cosx}}}$

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

Consider,

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

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

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

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

$\rm \: = \: \: {(secA + tanA)}^{2}$

$\rm \: = \: \: {\bigg(\dfrac{1}{cosA} + \dfrac{sinA}{cosA} \bigg) }^{2}$

$\rm \: = \: \: {\bigg(\dfrac{1 + sinA}{cosA} \bigg) }^{2}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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