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

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

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Alright, I need to solve the trigonometric problems presented. Let's tackle them one by one.

**Problem 1.2.1: Determine $$ \cos 52^{\circ} $$ in terms of $$ r $$, where $$ \cos 26^{\circ} = r $$.**

First, I recall the double-angle identity for cosine:
$$
\cos 2\theta = 2\cos^2 \theta - 1
$$
Given that $$ \cos 26^{\circ} = r $$, I can set $$ \theta = 13^{\circ} $$ because $$ 2\theta = 26^{\circ} $$. Plugging into the identity:
$$
r = 2\cos^2 13^{\circ} - 1
$$
Solving for $$ \cos^2 13^{\circ} $$:
$$
\cos^2 13^{\circ} = \frac{r + 1}{2}
$$
Now, I need to find $$ \cos 52^{\circ} $$. Notice that $$ 52^{\circ} = 2 \times 26^{\circ} $$, so I can use the double-angle identity again:
$$
\cos 52^{\circ} = 2\cos^2 26^{\circ} - 1
$$
But $$ \cos 26^{\circ} = r $$, so:
$$
\cos 52^{\circ} = 2r^2 - 1
$$
So, $$ \cos 52^{\circ} = 2r^2 - 1 $$.

**Problem 1.2.2: Determine $$ \tan 71^{\circ} $$ in terms of $$ r $$, where $$ \cos 26^{\circ} = r $$.**

First, recall that $$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$. To express $$ \tan 71^{\circ} $$ in terms of $$ r $$, I need to find $$ \sin 71^{\circ} $$ and $$ \cos 71^{\circ} $$ in terms of $$ r $$.

From the first problem, I have $$ \cos 52^{\circ} = 2r^2 - 1 $$. Now, $$ 71^{\circ} = 90^{\circ} - 19^{\circ} $$, so:
$$
\sin 71^{\circ} = \cos 19^{\circ}
$$
Using the double-angle identity for cosine:
$$
\cos 2\theta = 2\cos^2 \theta - 1
$$
Let $$ \theta = 19^{\circ} $$, so:
$$
\cos 38^{\circ} = 2\cos^2 19^{\circ} - 1
$$
But $$ \cos 38^{\circ} = \cos (90^{\circ} - 52^{\circ}) = \sin 52^{\circ} $$. From the first problem, $$ \cos 52^{\circ} = 2r^2 - 1 $$, so:
$$
\sin 52^{\circ} = \sqrt{1 - \cos^2 52^{\circ}} = \sqrt{1 - (2r^2 - 1)^2}
$$
This seems a bit convoluted. Maybe there's a simpler approach. Alternatively, I can use the identity:
$$
\tan \theta = \frac{1}{\cot \theta}
$$
But that might not directly help. Perhaps expressing $$ \tan 71^{\circ} $$ in terms of $$ \tan 19^{\circ} $$ would be more straightforward. However, without additional information, it's challenging to express $$ \tan 71^{\circ} $$ solely in terms of $$ r $$.

**Problem 1.3: Prove the identity $$ \frac{\sin 2x}{\cos 2x + \sin^2 x} = 2 \tan x $$.**

To prove this identity, I'll start by simplifying the left-hand side (LHS) and show that it equals the right-hand side (RHS).

First, recall the double-angle identities:
$$
\sin 2x = 2 \sin x \cos x
$$
$$
\cos 2x = \cos^2 x - \sin^2 x
$$
Substituting these into the LHS:
$$
\frac{\sin 2x}{\cos 2x + \sin^2 x} = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x + \sin^2 x} = \frac{2 \sin x \cos x}{\cos^2 x}
$$
Simplifying further:
$$
\frac{2 \sin x \cos x}{\cos^2 x} = \frac{2 \sin x}{\cos x} = 2 \tan x
$$
Which is exactly the RHS. Therefore, the identity is proven.

1.2.1. $$ \cos 52^{\circ} = 2r^2 - 1 $$ 1.2.2. $$ \tan 71^{\circ} $$ cannot be expressed solely in terms of $$ r $$ without additional information. 1.3. The identity $$ \frac{\sin 2x}{\cos 2x + \sin^2 x} = 2 \tan x $$ is proven as follows: $$ \frac{\sin 2x}{\cos 2x + \sin^2 x} = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x + \sin^2 x} = \frac{2 \sin x \cos x}{\cos^2 x} = 2 \tan x $$ Thus, the identity holds true.

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: \( \frac{\sin 2 x}{\cos 2 x+\sin ^{2} x}=2 \tan x \)

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