Đáp án:

 `a)` $\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right) \cdot \dfrac{1+\cos\left(\dfrac{\pi}{2}+x\right)}{\sin\left(\dfrac{\pi}{2}+x\right)} = 1$
->S.dụng các công thức lượng giác:
$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$
$\cos\left(\dfrac{\pi}{2}+x\right) = -\sin x$
$\sin\left(\dfrac{\pi}{2}+x\right) = \cos x$
$\tan\left(\dfrac{\pi}{4}\right) = 1$
`*`thừa số thứ nhất:
$\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right) = \dfrac{\tan\left(\dfrac{\pi}{4}\right) + \tan\left(\dfrac{x}{2}\right)}{1 - \tan\left(\dfrac{\pi}{4}\right)\tan\left(\dfrac{x}{2}\right)} = \dfrac{1 + \tan\left(\dfrac{x}{2}\right)}{1 - \tan\left(\dfrac{x}{2}\right)}$.
`*`thừa số thứ hai:
$\dfrac{1+\cos\left(\dfrac{\pi}{2}+x\right)}{\sin\left(\dfrac{\pi}{2}+x\right)} = \dfrac{1-\sin x}{\cos x}$.
Ta có: $\sin x = 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$ và $1 - \sin x = \sin^2\left(\dfrac{x}{2}\right) + \cos^2\left(\dfrac{x}{2}\right) - 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right) = \left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)^2$.
Và $\cos x = \cos^2\left(\dfrac{x}{2}\right) - \sin^2\left(\dfrac{x}{2}\right) = \left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)\left(\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)\right)$.
->Do đó, $\dfrac{1-\sin x}{\cos x} = \dfrac{\left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)^2}{\left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)\left(\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)\right)} = \dfrac{\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)}$.
-> chia cả tử và mẫu cho $\cos\left(\dfrac{x}{2}\right)$ (với điều kiện $\cos\left(\dfrac{x}{2}\right) \neq 0$):
$= \dfrac{\dfrac{\cos(x/2)}{\cos(x/2)} - \dfrac{\sin(x/2)}{\cos(x/2)}}{\dfrac{\cos(x/2)}{\cos(x/2)} + \dfrac{\sin(x/2)}{\cos(x/2)}} = \dfrac{1 - \tan\left(\dfrac{x}{2}\right)}{1 + \tan\left(\dfrac{x}{2}\right)}$.
`=>`
VT $= \dfrac{1 + \tan\left(\dfrac{x}{2}\right)}{1 - \tan\left(\dfrac{x}{2}\right)} \cdot \dfrac{1 - \tan\left(\dfrac{x}{2}\right)}{1 + \tan\left(\dfrac{x}{2}\right)} = 1 (đpcm)$.

`b)` $\tan\left(\dfrac{\pi}{4}+x\right) = \dfrac{1+\sin 2x}{\cos 2x}$}
Biến đổi vế phải (VP):
$VP = \dfrac{1+\sin 2x}{\cos 2x}$.
S/dụng công thức: $1 = \sin^2 x + \cos^2 x$ và $\sin 2x = 2\sin x \cos x$.
$\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.
$VP = \dfrac{\sin^2 x + \cos^2 x + 2\sin x \cos x}{(\cos x - \sin x)(\cos x + \sin x)}$.
$VP = \dfrac{(\sin x + \cos x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$.
$VP = \dfrac{\sin x + \cos x}{\cos x - \sin x}$.
->chia cả tử và mẫu cho $\cos x$ (với đ/kiện $\cos x \neq 0$):
$VP = \dfrac{\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\cos x}}{\dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x}} = \dfrac{\tan x + 1}{1 - \tan x}$.

Biến đổi vế trái (VT):
$VT = \tan\left(\dfrac{\pi}{4}+x\right) = \dfrac{\tan\left(\dfrac{\pi}{4}\right) + \tan x}{1 - \tan\left(\dfrac{\pi}{4}\right)\tan x} = \dfrac{1 + \tan x}{1 - \tan x} (đpcm)$.

`c)` $\cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right) = \dfrac{\cos x}{1-\sin x}$}
Biến đổi vế phải (VP):
$VP = \dfrac{\cos x}{1-\sin x}$.
`-` S.dụng công thức: $\cos x = \cos^2\left(\dfrac{x}{2}\right) - \sin^2\left(\dfrac{x}{2}\right) = \left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)\left(\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)\right)$.
$1 - \sin x = \sin^2\left(\dfrac{x}{2}\right) + \cos^2\left(\dfrac{x}{2}\right) - 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right) = \left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)^2$.
$VP = \dfrac{\left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)\left(\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)\right)}{\left(\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)\right)^2} = \dfrac{\cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right) - \sin\left(\dfrac{x}{2}\right)}$.
->chia cả tử và mẫu cho $\sin\left(\dfrac{x}{2}\right)$ (với điều kiện $\sin\left(\dfrac{x}{2}\right) \neq 0$):
$VP = \dfrac{\dfrac{\cos(x/2)}{\sin(x/2)} + \dfrac{\sin(x/2)}{\sin(x/2)}}{\dfrac{\cos(x/2)}{\sin(x/2)} - \dfrac{\sin(x/2)}{\sin(x/2)}} = \dfrac{\cot\left(\dfrac{x}{2}\right) + 1}{\cot\left(\dfrac{x}{2}\right) - 1}$.

Biến đổi vế trái (VT):
$VT = \cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$.
S/dụng công thức $\cot(A-B) = \dfrac{\cot A \cot B + 1}{\cot B - \cot A}$.
$VT = \dfrac{\cot\left(\dfrac{\pi}{4}\right)\cot\left(\dfrac{x}{2}\right) + 1}{\cot\left(\dfrac{x}{2}\right) - \cot\left(\dfrac{\pi}{4}\right)}$.
Vì $\cot\left(\dfrac{\pi}{4}\right) = 1$:
$VT = \dfrac{1 \cdot \cot\left(\dfrac{x}{2}\right) + 1}{\cot\left(\dfrac{x}{2}\right) - 1} = \dfrac{\cot\left(\dfrac{x}{2}\right) + 1}{\cot\left(\dfrac{x}{2}\right) - 1} (đpcm)$.