Question

$\pink{(cot A - cosec A )² = \frac{1- cos A}{1+ cos A}}(cotA−cosecA)²$
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Answer 1

$\large\underline{\sf{Given \: appropriate \:Question - }}$

Prove that

$\rm :\longmapsto\: {(cotA - cosecA)}^{2} = \dfrac{1 - cosA}{1 + cosA}$

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

Consider RHS

$\rm :\longmapsto\: {(cotA - cosecA)}^{2}$

can be rewritten as

$\rm \: = \: {\bigg( - (cosecA - cotA)\bigg) }^{2}$

$\rm \: = \: {(cosecA - cotA)}^{2}$

We know,

$\red{ \boxed{ \sf{ \:cotx = \frac{cosx}{sinx}}}} \: \: and \: \: \red{ \boxed{ \sf{ \:cosecx = \frac{1}{sinx}}}}$

$\rm \: = \: {\bigg[\dfrac{1}{sinA} - \dfrac{cosA}{sinA} \bigg]}^{2}$

$\rm \: = \: {\bigg[\dfrac{1 - cosA}{sinA}\bigg]}^{2}$

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

We know,

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

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

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

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

Hence,

$\rm :\longmapsto\: \red{ \boxed{ \sf{ \:{(cotA - cosecA)}^{2} = \dfrac{1 - cosA}{1 + cosA}}}}$

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