Question

PLEASE ANSWER THE QUE.

Answer 1

Answer:

Step-by-step explanation:

$\implies\frac{sinA}{1-cosA}\\\\\implies\frac{sinA}{1-cosA}\times\frac{1+cosA}{1+cosA}\\\\\implies\frac{sinA(1+cosA)}{(1-cosA)(1+cosA)}\\\\\implies\frac{sinA(1+cosA)}{1^2-cos^2A}}\\\\\implies\frac{sinA(1+cosA)}{1-cos^2A}\\\\\implies\frac{sinA(1+cosA)}{sin^2A}\\\\\implies\frac{1+cosA}{sinA}\\\\\implies\frac{1+cosA}{sinA}\\\\\implies\frac{1}{sinA}+\frac{cosA}{sinA}\\\\\implies cosecA+cotA$

Answer 2

Step-by-step explanation:

We know,

cosec²A − cot²A = 1

Using a² − b² = (a+b).(a−b)

(cosecA+cotA).(cosecA−cotA) = 1

cosecA − cotA = 1

cosecA + cotA

cosecA − cotA = 1

1 + cosA

sinA sinA

cosecA − cotA = 1

1+cosA

sinA

cosecA − cotA = sinA

1+cosA

Hence proved!!!