Với $0 < x< \dfrac{\pi}{2}$
Ta được: $\begin{cases}\sin x >0\\\cos x >0\end{cases}$
Khi đó:
\(\begin{array}{l}
\quad \sqrt{\cot^2x - \cos^2x} + \sqrt{\tan^2x - \sin^2x}\\
= \sqrt{\dfrac{\cos^2x}{\sin^2x} - \cos^2x} + \sqrt{\dfrac{\sin^2x}{\cos^2x} - \sin^2x}\\
= \sqrt{\dfrac{\cos^2x - \cos^2x\sin^2x}{\sin^2x}} + \sqrt{\dfrac{\sin^2x - \sin^2x\cos^2x}{\cos^2x}}\\
= \sqrt{\dfrac{\cos^2x(1 - \sin^2x)}{\sin^2x}} + \sqrt{\dfrac{\sin^2x(1 - \cos^2x)}{\cos^2x}}\\
= \sqrt{\dfrac{\cos^4}{\sin^2x}} + \sqrt{\dfrac{\sin^4x}{\cos^2x}}\\
= \left|\dfrac{\cos^2x}{\sin x}\right| + \left|\dfrac{\sin^2x}{\cos x}\right|\\
= \dfrac{\cos^2x}{\sin x} + \dfrac{\sin^2x}{\cos x}\\
= \dfrac{\cos^3x + \sin^3x}{\sin x\cos x}\\
= \dfrac{(\cos x + \sin x)(\cos^2x - \sin x\cos x + \sin^2x)}{\sin x\cos x}\\
= \dfrac{\cos x + \sin x}{\sin x\cos x}\cdot (\cos^2x - \sin x\cos x + \sin^2x)\\
= \left(\dfrac{1}{\sin x} + \dfrac{1}{\cos x}\right)(1 - \sin x\cos x)\qquad (đpcm)
\end{array}\)
