Question

rut gon

\(A=\frac{1-sinx-cos2x}{sin2x-cosx}\)

\(B=\frac{sin2x+sinx}{1+cos2x+cosx}\)

\(C=\frac{tana-cota}{tana+cota}+cos2a\)

Answer 1

\(A=\frac{1-sinx-1+2sin^2x}{2sinx.cosx-cosx}=\frac{sinx\left(2sinx-1\right)}{cosx\left(2sinx-1\right)}=\frac{sinx}{cosx}=tanx\)

\(B=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)

\(C=\frac{sina.cosa\left(tana-cota\right)}{sina.cosa\left(tana+cota\right)}+cos2a=\frac{sin^2a-cos^2a}{sin^2a+cos^2a}+cos2a\)

\(=-cos2a+cos2a=0\)