Question

Prove that
1+secA-tanA/1+secA+tanA=1-sinA/cosA​

Answer 1

LHS= (secA-tanA+1)/(secA+tanA+1)

We know the relation that,

1+tan^2A=sec^2A

=> 1=sec^2A-tan^2A

Applying the above relation in the LHS, we have,

LHS=(secA-tanA+sec^2A-tan^2A)/(secA+tanA+1)

=[secA-tanA+(secA-tanA)(secA+tanA)/(secA+tanA+1) [since, a^2 - b^2 = (a-b)(a+b)]

Taking (secA-tanA) common from the terms in the numerator

=(secA-tanA)(1+secA+tanA)/(secA+tanA+1)

= secA-tanA

= 1/cosA - sinA/cosA

=(1-sinA)/cosA

Hence, proved

Answer 2

Answer:

here is your answer.

Step-by-step explanation:

instead of 1 write sec square A - tan square A