Prove that cos 2A / 1+sin 2A=1-tan A / 1+tan A?

Answer 1

Hi,

1 - tan A
------------ =
1 + tan A


1 - sin A/cos A
---------------------
1 + sin A/cos A

Multiply by cos A on top and bottom.

cos A - sin A
------------------
cos A + sin A

Multiply top and bottom both by (cos A + sin A)

(cos A - sin A)(cos A + sin A)
------------------------------------------ =
(cos A + sin A)(cos A + sin A)

cos² A - sin² A
------------------------------------------ =
cos² A + 2sin A cos A + sin² A

cos 2A
------------------------------------------- =
(cos² A + sin² A) + 2sin A cos A


cos 2A
------------- <== answer (left hand side)
1+ sin 2A

I hope that helps!! :-)

Answer 2

taking the rhs

write as sin and cos


so 1 - s/c / 1 + s/c

now times by c

c - s / c + s


now multiply top and bottom by (c + s)

c^2 - s^2 / (c + s)^2

top line using double angle formula becomes cos2A

expland bottom line = c^2 + s^2 + 2sc

but c^2 + s^2 = 1 and 2sc = sin 2A

so = cos2A/1-sin2A

Answer 3

Recall the following basic formulas:

(1) cos2A = cos²A - sin²A,

(2) sin2A = 2sinA*cosA, in terms of half the angles, and

(3) cos²A + sin²A = 1.

Now write down: cos2A/(1 + sin2A)

= (cos²A - sin²A)/(cos²A + sin²A + 2sinA*cosA)

= [(cosA + sinA)(cosA - sinA)]/(cosA + sinA)²

= (cosA - sinA)/(cosA + sinA)

= (1 - tanA)/(1 + tanA) after dividing the numerator and denominator both by cosA ≠ 0.