Question

10, Prove that
(tan A + cosec B)2 - (cot B - sec A)2 = 2 tan A cosec B + 2 cot B sec A​

Answer 1

Step-by-step explanation:

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