Question

Prove that CosecA ( 1- CosA) (CosecA + CotA ) = 1​

Answer 1

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

Consider,

$\rm :\longmapsto\:cosecA(1 - cosA)(cosecA + cotA)$

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

$\: \: \: \: \red{ \sf \because \: \boxed{ \red{ \sf \: cosecx = \dfrac{1}{sinx}}}} \: \: \: \: \: \: \: \: \red{\boxed{\red{ \sf \: cotx = \dfrac{cosx}{sinx}}}}$

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

$\:  \: \rm  =  \:  \: \dfrac{1 - {cos}^{2} A}{ {sin}^{2}A}$

$\: \: \: \: \: \: \: \: \: \because \: \red{\boxed{ \red{ \sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

$\:  \: \rm  =  \:  \: \dfrac{{sin}^{2} A}{ {sin}^{2}A}$

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

$\:  \: \rm  =  \:  \: 1$

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

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

Answer 2

Step-by-step explanation:

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

Consider,

$\rm :\longmapsto\:cosecA(1 - cosA)(cosecA + cotA)$

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

$\: \: \: \: \red{ \sf \because \: \boxed{ \red{ \sf \: cosecx = \dfrac{1}{sinx}}}} \: \: \: \: \: \: \: \: \red{\boxed{\red{ \sf \: cotx = \dfrac{cosx}{sinx}}}}$

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

$\:  \: \rm  =  \:  \: \dfrac{1 - {cos}^{2} A}{ {sin}^{2}A}$

$\: \: \: \: \: \: \: \: \: \because \: \red{\boxed{ \red{ \sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

$\:  \: \rm  =  \:  \: \dfrac{{sin}^{2} A}{ {sin}^{2}A}$

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

$\:  \: \rm  =  \:  \: 1$

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

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