Question

prove how $(cosA/1-sinA) - (sinA/1-cosA)=secA + cosecA +secAcosecA$

Answer 1

Appropriate Question :-

Prove that :-

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

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

Consider LHS

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

On rationalizing each term, we get

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

We know that,

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

So, using this, we get

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

We know that,

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

So, using this we get

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

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

$\rm \: = \: \dfrac{sinA + {sin}^{2}A + cosA + {cos}^{2}A}{cosA \: sinA}$

$\rm \: = \: \dfrac{sinA + cosA + {sin}^{2}A + {cos}^{2}A}{cosA \: sinA}$

We know,

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

So, using this, we get

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

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

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

$\rm \: = \: secA + cosecA + secA \: cosecA$

Hence,

$\boxed{ \tt{ \: \dfrac{cosA}{1 - sinA} + \dfrac{sinA}{1 - cosA} =secA + cosecA+ secAcosecA}}$

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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:

pls check the Question ... I have attached the explanation