Đáp án + Giải thích các bước giải:

$\textbf{b}\bigg)$

$\begin{array}{l}\textbf{VT} &= \dfrac{\tan \alpha + 1}{\tan \alpha - 1} \bigg(x \ne \dfrac{\pi}{4} + k\pi; x \ne \dfrac{k\pi}{2}; k \in \mathbb{Z}\bigg)
\\ &= \dfrac{\tan \alpha \big(1 + \frac{1}{\tan \alpha}\big)}{\tan \alpha\big(1 - \frac{1}{\tan \alpha}\big)}
\\ &= \dfrac{1 + \cot \alpha}{1 - \cot \alpha} = \textbf{VP}\end{array}$

$\textbf{c}\bigg)$

$\begin{array}{l}\textbf{VT} &= \tan^2 \alpha - \sin^2 \alpha \bigg(x \ne \dfrac{\pi}{2} + k\pi; k \in \mathbb{Z})
\\ &= \sin^2 \alpha \bigg(\dfrac{1}{\cos^2 \alpha} - 1\bigg)
\\ &= \sin^2 \alpha \bigg(\dfrac{1 - \cos^2 \alpha}{\cos^2 \alpha}
\\ &= \sin^2 \alpha \cdot \dfrac{\sin^2 \alpha}{\cos^2 \alpha}
\\ &= \sin^2 \alpha \tan^2 \alpha = \textbf{VP}\end{array}$

$\textbf{d}\bigg)$

$\begin{array}{l} \textbf{VT} &= \dfrac{1 - 4\sin^2 \alpha \cos^2 \alpha}{(\sin \alpha - \cos \alpha)^2} \bigg(x \ne \dfrac{\pi}{4} + k\pi; k \in \mathbb{Z}\bigg)
\\ &= \dfrac{(1 + 2\sin \alpha \cos \alpha)(1 - 2\sin \alpha \cos \alpha)}{(\sin \alpha - \cos \alpha)^2}
\\ &= \dfrac{(\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha)(\sin^2 \alpha - 2\sin\alpha \cos \alpha + \cos^2 \alpha)}{(\sin \alpha - \cos \alpha)^2}
\\ &= \dfrac{(\sin \alpha + \cos \alpha)^2(\sin \alpha - \cos \alpha)^2}{(\sin \alpha - \cos \alpha)^2}
\\ &= (\sin \alpha + \cos \alpha)^2 =\textbf{VP} \end{array}$

$\textbf{e}\bigg)$

$\begin{array}{l} \textbf{VT} &= \dfrac{1 - 2\cos^2 \alpha}{1 + 2\sin \alpha \cos \alpha} \bigg(x \ne -\dfrac{\pi}{4} + k\pi; k \in \mathbb{Z}\bigg)
\\ &= \dfrac{\sin^2 \alpha + \cos^2 \alpha - 2\cos^2 \alpha}{\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha}
\\ &= \dfrac{\sin^2 \alpha - \cos^2 \alpha}{(\sin \alpha + \cos \alpha)^2}
\\ &= \dfrac{(\sin \alpha + \cos \alpha)(\sin \alpha - \cos \alpha)}{(\sin \alpha + \cos \alpha)^2}
\\ &= \dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} = \textbf{VP} \end{array}$