Question

(cosec B + cot B ) ( 1 - cos B) =
please answer ​

Answer 1

Answer:

Answer

We have,

LHS = (tanA+cosecB)

2

−(cotB−secA)

2

⇒ LHS = (tan

2

A + cosec

2

B +2tanA⋅cosecB)−(cot

2

B + sec

2

A - 2cotB⋅secA)

⇒ LHS = (tan

2

A - sec

2

A )+(cosec

2

B - cot

2

B)+2tanA⋅cosecB+2cotB⋅secA

⇒ LHS =−1+1+2tanA⋅cosecB+2cotB⋅secA

⇒ LHS =2(tanA⋅cosecB+cotB⋅secA)

⇒ LHS = 2 tan A cot B (

cotB

cosecB

+

tanA

secA

) [Dividing and multiplying by tan A cot B]

⇒ LHS = 2 tan A cot B

⎩

⎪

⎪

⎨

⎪

⎪

⎧

sinB

cosB

sinB

1

+

ocsA

sinA

cosA

1

⎭

⎪

⎪

⎬

⎪

⎪

⎫

⇒ LHS = 2 tan A cot B (

cosB

1

+

sinA

1

) =2tanAcotB(secB+cosecA)= RHS

Answer 2

Answer:

SIMPLIFY;:

$\frac{cosec.b + sin(b))cos(b) + cosec.b}{cos(b) + 1}$

FACTOR;:

$cosec.b - cot(b)cos(b) + cot(b)$

this is a answer