Answer :
[tex]A=x\\\\(secx+tanx)^2=\frac{1+sinx}{1-sinx}\\\\L=\left(\frac{1}{cosx}+\frac{sinx}{cosx}\right)^2=\left(\frac{1+sinx}{cosx}\right)^2=\frac{(1+sinx)^2}{cos^2x}\\\\=\frac{(1+sinx)^2}{1-sin^2x}=\frac{(1+sinx)^2}{1^2-sin^2x}=\frac{(1+sinx)^2}{(1-sinx)(1+sinx)}=\frac{1+sinx}{1-sinx}=R[/tex]
[tex]if\ R=\frac{1+sinx}{\frac{1}{sinx}}=(1+sinx)sinx=sinx+sin^2x\ then\ L\neq R[/tex]
[tex]secx=\frac{1}{cosx}\\\\tanx=\frac{sinx}{cosx}\\\\sin^2x+cos^2x=1\to cos^2x=1-sin^2x\\\\a^2-b^2=(a-b)(a+b)[/tex]
[tex]if\ R=\frac{1+sinx}{\frac{1}{sinx}}=(1+sinx)sinx=sinx+sin^2x\ then\ L\neq R[/tex]
[tex]secx=\frac{1}{cosx}\\\\tanx=\frac{sinx}{cosx}\\\\sin^2x+cos^2x=1\to cos^2x=1-sin^2x\\\\a^2-b^2=(a-b)(a+b)[/tex]