<s>[INST] solving a trigonometric identity: $\frac{\sec^2 x + 2 \tan x}{1 + \tan x}=1+\tan x$. $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = 1 +\tan x$$    I started by making $1/\cos^2 x$ the
$2\tan x/\cos x$ making them but I can't came the the right answer.
 [/INST] We use the identity $$\sec^2 x = 1 + \tan^2 x$$
This gives us $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = \dfrac{\overbrace{(1 + \tan^2 x )+ 2\tan x}^{(1 + \tan x)^2}}{1 + \tan x} = \dfrac{(1 + \tan x)^2}{1 + \tan x} = 1 +\tan x$$
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