Question

[Đông Viên]

cm các đẳng thức:

a) \(\frac{1+\sin^2\alpha}{1-\sin^2\alpha}=1+2\tan^2\alpha\)

b) \(\frac{\cos\alpha}{1+\sin\alpha}+\tan\alpha=\frac{1}{\cos\alpha}\)

c) \(\frac{\sin\alpha}{1+\cos\alpha}+\frac{1+\cos\alpha}{\sin\alpha}=\frac{2}{\sin\alpha}\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)

\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)

\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)