Đáp án:
Có thể lấy VT-VP rồi chứng minh chúng bằng 0 thì VT-VP.
$\begin{array}{l}
VT - VP\\
= {\sin ^2}x\left( {1 + \cot x} \right) + {\cos ^2}x\left( {1 + {\mathop{\rm tanx}\nolimits} } \right)\\
- {\left( {\sin x + \cos x} \right)^2}\\
= {\sin ^2}x\left( {1 + \dfrac{{\cos x}}{{\sin x}}} \right) + {\cos ^2}x\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)\\
- {\left( {\sin x + \cos x} \right)^2}\\
= {\sin ^2}x.\dfrac{{\sin x + \cos x}}{{\sin x}} + {\cos ^2}x.\dfrac{{\cos x + \sin x}}{{\cos x}}\\
- {\left( {\sin x + \cos x} \right)^2}\\
= \sin x.\left( {\sin x + \cos x} \right) + \cos x.\left( {\sin x + \cos x} \right)\\
- {\left( {\sin x + \cos x} \right)^2}\\
= \left( {\sin x + \cos x} \right)\left( {\sin x + \cos x} \right) - {\left( {\sin x + \cos x} \right)^2}\\
= {\left( {\sin x + \cos x} \right)^2} - {\left( {\sin x + \cos x} \right)^2}\\
= 0\\
Vậy\,VT - VP = 0 \Rightarrow VT = VP
\end{array}$
