Question

1/(secA+tanA)-1/cosA=1/cos-1/(secA-tanA)
you have to prove this.​

Answer 1

Answer:

 

To prove this I am taking LHS first

Take RHS now

LHS = RHS

Hope it helps you

Step-by-step explanation:

Answer 2

Taking R.H.S.

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

We know that, 1/cosA = secA and rationalize 1/(secA - tanA)

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

Used identity: (a + b)(a - b) = a² - b²

= secA - (secA + tanA)/(sec²A - tan²A)

= secA - (secA + tanA)

= secA - secA - tanA

=  - tanA

Taking L.H.S.

= 1/(secA + tanA) - 1/cosA

We know that, 1/cosA = secA and rationalize 1/(secA + tanA)

= (secA-tanA)/[(secA + tanA)(secA - tanA)] - secA

Used identity: (a + b)(a - b) = a² - b²

= (secA - tanA)/(sec²A - tan²A) -secA

= secA - tanA - secA

= -  tanA

L.H.S = R.H.S.

Hence, proved.