Verified Answer :

Ryan2

\(sinx.tanx=secx-cosx\\ \\ sinx.\frac{sinx}{cosx}=\frac{1}{cosx}-cosx\\ \\ \frac{sin^2x}{cosx}=\frac{1-cos^2x}{cosx}\\ \\ \frac{sin^2x}{cosx}=\frac{sin^2x}{cosx}\)

Related Questions