Can someone please explain (cos(x)+sin(x))/(cos(x)-sin(x)=sec(2x)+tan(2x). I have no luck with this. Thanks

RS

= 1/cos2x + sin2x/cos2x
= 1/(cos^2x - sin^2x) + 2sinxcosx/cos^2x - sin^2x)
= (1 + 2sinxcosx)/[(cosx+sinx)(cosx-sinx)]
= (sin^2x + 2sinxcosx + cos^2x)/[(cosx+sinx)(cosx-sinx)]
= (sinx + cosx)^2/[(cosx+sinx)(cosx-sinx)]
= (sinx + cosx)/(cosx - sinx)
= LS

Thank you very much!

Certainly! To prove the equation (cos(x) + sin(x))/(cos(x) - sin(x)) = sec(2x) + tan(2x), we need to use trigonometric identities and algebraic manipulations.

Step 1: Start with the left-hand side (LHS).

(cos(x) + sin(x))/(cos(x) - sin(x))

Step 2: To simplify this expression, we can multiply both the numerator and denominator by the conjugate of the denominator, which is (cos(x) + sin(x)).

[(cos(x) + sin(x)) * (cos(x) + sin(x))]/[(cos(x) - sin(x)) * (cos(x) + sin(x))]

Step 3: Expand the numerator and denominator.

[cos²(x) + 2cos(x)sin(x) + sin²(x)]/[cos²(x) - sin²(x)]

Step 4: Simplify the numerator.

[1 + 2cos(x)sin(x)]/[cos²(x) - sin²(x)]

Step 5: Recall the trigonometric identity: sin(2x) = 2sin(x)cos(x) and cos(2x) = cos²(x) - sin²(x).

[1 + sin(2x)]/[cos(2x)]

Step 6: Recall the trigonometric identity: sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x).

[sec(2x) + sin(2x) * tan(2x)]/[cos(2x)]

Step 7: Distribute tan(2x) term.

[sec(2x) + tan(2x) * sin(2x)]/[cos(2x)]

Step 8: Recognize that sin(2x)/cos(2x) is equal to tan(2x).

[sec(2x) + tan(2x)]/[cos(2x)]

Step 9: Since the numerator and denominator are the same as the right-hand side (RHS), we have proven that

(cos(x) + sin(x))/(cos(x) - sin(x)) = sec(2x) + tan(2x).

This completes the explanation and proof of the given equation.