Question

Prove the following identity:<br />cos A /1+ sin A<br />+<br />1+ sin A/ cos A<br />= 2 sec A<br />​

Answer 1

$\large\underline{\bold{Given\:Question - }}$

$\tt \: To \: Prove : \: \dfrac{1 + sinA}{cosA} + \dfrac{cosA}{1 + sinA} = 2 \: secA$

$\large\underline{\bold{Solution :- }}$

$\large \underline{\tt \:{Identities \: used }}$

$(1). \: \: \boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }$

$(2). \: \: \boxed{ \bf \: {sin}^{2}A + {cos}^{2}A = 1}$

$(3). \: \: \boxed{ \bf \: secA = \dfrac{1}{cosA} }$

$\rm :\longmapsto\: \bf \: Consider \: LHS$

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

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

$\rm :\longmapsto\:\dfrac{1 + {sin}^{2}A + 2sinA + {cos}^{2}A }{(1 +sinA)cosA }$

$\rm :\longmapsto\:\dfrac{1 + {sin}^{2}A + {cos}^{2}A + 2sinA }{(1 + sinA)cosA}$

$\rm :\longmapsto\:\dfrac{1 + 1 + 2sinA}{(1 + sinA)cosA}$

$\rm :\longmapsto\:\dfrac{2 + 2sinA}{(1 + sinA)cosA}$

$\rm :\longmapsto\:\dfrac{2 \: \cancel{(1 + sinA)}}{ \cancel{(1 + sinA)} \: cosA}$

$\rm :\longmapsto\:\dfrac{2}{cosA}$

$\rm :\longmapsto\:2 \: secA$

$\rm :\implies\:RHS$

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

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