g: \(\dfrac{2}{tanx-1}+\dfrac{cotx+1}{cotx-1}\)
\(=\dfrac{2}{\dfrac{1}{cotx}-1}+\dfrac{cotx+1}{cotx-1}\)
\(=2:\dfrac{1-cotx}{cotx}+\dfrac{cotx+1}{cotx-1}\)
\(=\dfrac{-2cotx}{cotx-1}+\dfrac{cotx+1}{cotx-1}=\dfrac{-cotx+1}{cotx-1}=-1\)(ĐPCM)
h: \(\sqrt{sin^4x+4\cdot cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)
\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{sin^4x-4\cdot sin^2x+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)
\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)
\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)
\(0< =sin^2x< =1\)
=>\(-2< =sin^2x-2< =-1\)
=>\(\left|sin^2x-2\right|=2-sin^2x\)
\(0< =cos^2x< =1\)
=>\(-2< =cos^2x-2< =-1\)
=>\(\left|cos^2x-2\right|=2-cos^2x\)
Do đó: \(\sqrt{sin^4x+4\cdot cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}=2-sin^2x+2-cos^2x=4-1=3\)
=>ĐPCM
i: \(\dfrac{tanx+tany}{cotx+coty}=\dfrac{tanx+tany}{\dfrac{1}{tanx}+\dfrac{1}{tany}}\)
\(=\left(tanx+tany\right):\dfrac{tanx+tany}{tanx\cdot tany}=tanx\cdot tany\)(ĐPCM)
k: \(\dfrac{tan^2x-tan^2y}{tan^2x\cdot tan^2y}=\dfrac{1}{tan^2y}-\dfrac{1}{tan^2x}\)
\(\dfrac{sin^2x-sin^2y}{sin^2x\cdot sin^2y}=\dfrac{1}{sin^2y}-\dfrac{1}{sin^2x}\)
\(=\left(1+cot^2y\right)-\left(1-cot^2x\right)=cot^2y-cot^2x\)
\(=\dfrac{1}{tan^2y}-\dfrac{1}{tan^2x}\)
Do đó: \(\dfrac{sin^2x-sin^2y}{sin^2x\cdot sin^2y}=\dfrac{tan^2x-tan^2y}{tan^2x\cdot tan^2y}\)
