Question

Chứng minh các đẳng thức sau (giả sử các biểu thức đã cho đều có nghĩa

1)\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2013}=\frac{sinx^{2013}+cot^{2013}}{1+sin^{2013}.tanx^{2013}}\)

2) \(\left(\sqrt{\frac{1+sinx}{1-sinx}}-\sqrt{\frac{1-sin}{1+sinx}}\right)^2=4tan^2x\)

Answer 1

a/ \(\frac{sinx+cotx}{1+sinx.tanx}=\frac{sinx.cosx\left(sinx+cotx\right)}{sinx.cosx\left(1+sinx.tanx\right)}=\frac{cosx\left(sin^2x+cosx\right)}{sinx\left(cosx+sin^2x\right)}=cotx\)

\(\Rightarrow VT=cot^{2013}x\)

\(VP=\frac{sin^{2013}x+cot^{2013}x}{1+sin^{2013}x.tan^{2013}x}=\frac{sin^{2013}x.cos^{2013}x\left(sin^{2013}x+cot^{2013}x\right)}{sin^{2013}x.cos^{2013}x\left(1+sin^{2013}x.tan^{2013}x\right)}\)

\(=\frac{cos^{2013}x\left(sin^{4026}x+cos^{2013}x\right)}{sin^{2013}x\left(cos^{2013}x+sin^{4036}x\right)}=\frac{cos^{2013}x}{sin^{2013}x}=cot^{2013}x=VT\) (đpcm)

b/ \(\left(\sqrt{\frac{1+sinx}{1-sinx}}-\sqrt{\frac{1-sinx}{1+sinx}}\right)^2=\frac{1+sinx}{1-sinx}+\frac{1-sinx}{1+sinx}-2\)

\(=\frac{\left(1+sinx\right)^2+\left(1-sinx\right)^2}{\left(1-sinx\right)\left(1+sinx\right)}-2=\frac{2+2sin^2x}{1-sin^2x}-2=\frac{2+2sin^2x}{cos^2x}-2\)

\(=\frac{2}{cos^2x}-2+2tan^2x=\frac{2\left(1-cos^2x\right)}{cos^2x}+2tan^2x=2tan^2x+2tan^2x=4tan^2x\)