a)Giả sử điều đã cho đúng:
$\frac{tan2\alpha +cot3\beta }{tan3\beta +cot2\alpha }=\frac{tan2\alpha }{tan3\beta }$
$\Leftrightarrow \tan 2\alpha \tan 3\beta +1=1+\tan 2\alpha \tan 3\beta$
$\Leftrightarrow \tan 2\alpha \tan 3\beta=\tan 2\alpha \tan 3\beta$
điều này hiển nhiên đúng, nên đẳng thức đã cho cũng đúng.
b/$(1+\frac{1-cos\alpha}{1+cos\alpha})(1+\frac{1+cos\alpha}{1-cos\alpha})=\frac{4}{sin^2\alpha}$
$\frac{2}{1+\cos \alpha }.\frac{2}{1-\cos \alpha }=\frac{4}{1-\cos ^{2}\alpha }=\frac{4}{\sin ^{2}\alpha }$
c/$\frac{sin^4x+cos^4x-1}{sin^6x+cos^6x-1}=\frac{(\sin ^{2}x+\cos ^{2}x)^{2}-2\sin ^{2}x\cos ^{2}x-1}{(\sin ^{2}x+\cos ^{2}x)(1-3\sin ^{2}x\cos ^{2}x)-1}=\frac{-2\sin ^{2}x\cos ^{2}x}{-3\sin ^{2}x\cos ^{2}x}=\frac{2}{3}$
d/$(\frac{\sqrt{tan\alpha}+\sqrt{cot\alpha}}{sin\alpha+cos\alpha})^2=\frac{\tan \alpha +\cot \alpha +2\sqrt{\tan \alpha \cot \alpha }}{1+2\sin \alpha \cos \alpha }=\frac{\tan \alpha +\cot \alpha +2}{1+2\sin \alpha \cos \alpha }$
$\frac{\frac{\sin ^{2}\alpha +\cos ^{2}\alpha }{\sin \alpha \cos \alpha }+2}{1+2\sin \alpha \cos \alpha }=\frac{\frac{1+2\sin \alpha \cos \alpha}{\sin \alpha \cos \alpha}}{1+2\sin \alpha \cos \alpha}=\frac{1}{\sin \alpha \cos \alpha}$