Question

SOLVE:-


cos A - sin A + 1 / cos A + sin A -1 = cosec A + cot A , using the  identity  cosec ² A = 1+ cos² A​

Answer 1

Appropriate Question :-

Prove that,

$\sf \: \dfrac{cosA - sinA + 1}{cosA + sinA - 1} = cosecA + cotA, \: using \: {cosec}^{2}A = 1 + {cot}^{2}A$

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

Consider, LHS

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

On dividing numerator and denominator by sinA, we get

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

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

We know, that

$\boxed{ \tt{ \: \frac{cosx}{sinx} = cotx \: }}$

So, using this, we get

$\sf \: = \: \dfrac{cotA - 1 + cosecA}{cotA + 1 - cosecA}$

can be re-arranged as

$\sf \: = \: \dfrac{cotA+ cosecA - 1}{cotA + 1 - cosecA}$

Now, given that, using

$\boxed{ \tt{ \: {cosec}^{2}x = 1 + {cot}^{2}x\rm \implies\: {cosec}^{2}x - {cot}^{2}x = 1}}$

So, in numerator replace 1, we get

$\sf \: = \: \dfrac{cotA+ cosecA - ( {cosec}^{2}A - {cot}^{2}A)}{cotA + 1 - cosecA}$

We know,

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

So, using this identity in numerator, we get

$\sf \: = \: \dfrac{cotA + cosecA - (cosecA + cotA)(cosecA - cotA)}{cotA + 1 - cosecA}$

$\sf \: = \: \dfrac{(cotA + cosecA) \: [1 - (cosecA - cotA) \: ]}{cotA + 1 - cosecA}$

$\sf \: = \: \dfrac{(cotA + cosecA) \: \cancel{ [1 - cosecA + cotA\: ]}}{ \cancel{cotA + 1 - cosecA}}$

$\sf \: = \: cotA + cosecA$

Hence,

$\: \: \: \red{\boxed{ \tt{ \: \: \dfrac{cotA - 1 + cosecA}{cotA + 1 - cosecA} = cosecA + cotA \: }}}$

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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

Step-by-step explanation:

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