Question

anybody prove it please​

Answer 1

Step-by-step explanation:

Given : tanA=2

∴cotA=

tanA

1

=

2

1

∵1+cot

2

A=cosec

2

A

⇒cosec

2

A=1+(

2

1

)

2

=1+

4

1

=

4

5

⇒cosecA=

4

5

=

2

5

Now secAsinA+tan

2

A−cosecA

⇒

cosA

1

sinA+tan

2

A−cosecA∵secθ=

cosθ

1

⇒tanA+tan

2

A−cosecA ∵tanθ=

cosθ

sinθ

⇒2+2

2

−

2

5

⇒2+4−

2

5

⇒6−

2

5

=

2

12−

5

∴ The value of secAsinA+tan

2

A−cosecA=

2

12−

5

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Answer 2

Answer:

TanA[(SecA+1+SecA-1)/(SecA-1)(secA+1)]=0

=TanA(secA)/secA-1= TanA(SecA)/TanA

=SecA/TanA=cosecA

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