Question

prove that √secA-tanA/secA +tanA= 1-sinA/cosA​

Answer 1

Answer:

Step-by-step explanation:

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