Giải thích các bước giải:
$a)VT=\dfrac{1}{1+\tan\alpha}+\dfrac{1}{1+\cot\alpha}\\ =\dfrac{1}{1+\dfrac{\sin\alpha}{\cos\alpha}}+\dfrac{1}{1+\dfrac{\cos\alpha}{\sin\alpha}}\\ =\dfrac{1}{\dfrac{\sin\alpha+\cos\alpha}{\cos\alpha}}+\dfrac{1}{\dfrac{\cos\alpha+\sin\alpha}{\sin\alpha}}\\ =\dfrac{\cos\alpha}{\sin\alpha+\cos\alpha}+\dfrac{\sin\alpha}{\cos\alpha+\sin\alpha}\\ =\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha+\cos\alpha}\\ =1\\ =VP\\ b)VT=\sin^4x-\cos^4x\\ =(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)\\ =\sin^2x-\cos^2x\\ =\sin^2x-(1-\sin^2x)\\ =2\sin^2x-1\\ =VP\\ c)VT=\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}\\ =\dfrac{\sin^2x+\cos^2x}{\sin^2x}+\dfrac{\sin^2x+\cos^2x}{\cos^2x}\\ =1+\dfrac{\cos^2x}{\sin^2x}+1+\dfrac{\sin^2x}{\cos^2x}\\ =2+\cot^2x+\tan^2x\\ =VP\\ d)VT=\dfrac{1+\sin^2\alpha}{1-\sin^2\alpha}\\ =\dfrac{1+\sin^2\alpha}{\cos^2\alpha}\\ =\dfrac{1}{\cos^2\alpha}+\dfrac{\sin^2\alpha}{\cos^2\alpha}\\ =\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}+\tan^2\alpha\\ =\tan^2\alpha+1+\tan^2\alpha\\ =1+2\tan^2\alpha\\ =VP$
