Question

Prove that sin A/1-cos A=cosec A+ cot A​

Answer 1

Step-by-step explanation:

= sin A/1-cos A

multiply and divide by 1 + cosA

therefore

sinA(1 + cosA)/(1+cosA)(1-cosA)

sinA(1 + cosA)/(1-cossquareA)

now sinsquareA = 1-cossquareA

therefore

sinA(1 + cosA)/(sinsquareA)

now, (1 + cosA)/(sinA)

1/ sinA= cosecA and cosA/sinA= cotA

therefore cosecA+ cotA

hence prove

Answer 2

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

$\sf \: Prove \: that : \: \dfrac{sinA}{1 - cosA} = cosecA + cotA$

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

Identities Used :-

$\boxed{ \sf \: 1 - {cos}^{2}A = {sin}^{2}A}$

$\boxed{ \sf \: \dfrac{1}{sinA} = cosecA}$

$\boxed{ \sf \: \dfrac{cosA}{sinA} = cotA}$

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

Let's solve the problem now!!

Consider,

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

On multiply and divide by 1 + cosA, we get

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

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

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

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

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

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

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

─━─━─━─━─━─━─━─━─━─━─━─━─

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