1.2 If \( \cos 26=r \), determine the following in terms of \( r \), in its simplest form: \( 1.2 .1 \cos 52^{\circ} \) \( 1.2 .2 \tan 71^{\circ} \) 1.3 Prove the identity: \( \frac{\sin 2 x}{\cos 2 x+\sin ^{2} x}=2 \tan x \)

To find \( \cos 52^\circ \) in terms of \( r \) where \( \cos 26^\circ = r \), we use the co-function identity. Since \( \cos 52^\circ = \sin(90^\circ - 52^\circ) = \sin 38^\circ \) and using the sine double angle identity, \( \sin 38^\circ = \sin(2 \times 19) = 2 \sin 19 \cos 19 \). Now, by further breaking down \( \sin 19^\circ \) and \( \cos 19^\circ \), you can express them in terms of \( r \) through various trigonometric identities ultimately leading to an expression solely in terms of \( r \). For \( \tan 71^\circ \), we can use the identity \( \tan (90^\circ - x) = \cot x \). Thus, \( \tan 71^\circ = \cot 19^\circ \). From the cotangent definition, we know \( \cot x = \frac{\cos x}{\sin x} \). Therefore, express \( \cos 19^\circ \) and \( \sin 19^\circ \) in terms of \( r \) to get the required result. Now, for the identity \( \frac{\sin 2x}{\cos 2x + \sin^2 x} = 2 \tan x \), we start by rewriting \( \sin 2x \) as \( 2 \sin x \cos x \) and \( \cos 2x \) as \( \cos^2 x - \sin^2 x \). Substitute these into the left-hand side which simplifies into \( 2 \tan x \) upon recognizing common terms and simplifying the fraction. This should confirm the desired identity.