Question

1+tanA/sinA+1+cotA/cosA=2(secA+cosecA)?

Answer 1

This should read: Show

`{ 1 + tan A}/{sin A} + {1 + cot A}/{cos A} = 2(sec A + csc A)`

Explanation:

I'll assume this is a problem to prove, and should read

Show `{ 1 + tan A}/{sin A} + {1 + cot A}/{cos A} = 2(sec A + csc A)`

Let's just get the common denominator and add and see what happens.

`{ 1 + tan A}/{sin A} + {1 + cot A}/{cos A}`

`= {cos A(1 + sin A/ cos A) + sin A(1+cos A/sin A)}/{sin A cos A}`

`= {cos A + sin A + sin A + cos A}/{sin A cos A}`

`= {2cos A }/{sin A cos A} + {2 sin A}/{sin A cos A}`

`= 2( 1/sin A + 1/cos A)`

`= 2(csc A + sec A)`

`= 2(sec A + csc A) quad sqrt`

Answer 2

Verified below

Explanation:

`(1+tanA)/sinA+(1+cotA)/cosA=2(secA+cscA)`

Split the numerator:
`1/sinA+tanA/sinA+1/cosA+cotA/cosA=2(secA+cscA)`

Apply the reciprocal identities: `1/sinA= cscA`, `1/cosA= secA`:
`cscA+tanA/sinA+secA+cotA/cosA=2(secA+cscA)`

Apply the quotient identities: `cotA= cosA/sinA`, `tanA=sinA/cosA`:
`cscA+cancel(sinA)/(cosA/cancel(sinA))+secA+cancel(cosA)/(sinA/cancel(cosA))=2(secA+cscA)`

Apply the reciprocal identities:
`cscA+secA+secA+cscA=2(secA+cscA)`

Combine like terms:
`2cscA+2secA=2(secA+cscA)`

Factor out the 2:
`2(secA+cscA)= 2(secA+cscA)`