Question

prove that (1-tanA)(cosA/cos2A)=1/cosA+sinA

Answer 1

Step-by-step explanation:

LHS : (1 - tanA ) ( cos A / cos 2A )

cos A / cos 2A - sinA / cos 2A

( cos A - sinA ) / cos 2A

( cos A - sinA ) / ( 2 cos^2 A - 1 )

( cos A - sinA ) / (2 cos^2 A - sin^2 A - cos^2 A )

( cos A - sinA ) / (cos^2A - sin^2 A)

( cos A - sinA ) / ( cos A - sinA )( cos A + sinA )

1 / ( cos A + sinA )

= RHS

Answer 2

As Cos(2A) = [{Cos(A)}*(2) - {Sin(A)}*(2)]
& also a*2 - b*2 = (a-b)(a+b)

LHS ->
=> [1-{sinA/cosA}] [cosA/(cosA-sinA)(cosA+sinA)]
=> [ {cosA-sinA}/cosA ] [ cosA/(cosA-sinA)(cosA+sinA)]
By cutting cosA-sinA & cosA both in num and den we get;
=>1/cosA+sinA
Hence proved
LHS = RHS