Question Number 129051 by oustmuchiya@gmail.com last updated on 12/Jan/21
$${prove}\:{that}\:\frac{\mathrm{1}+\boldsymbol{{sinx}}}{\boldsymbol{{sinxcosx}}}= \\ $$$$\boldsymbol{{tanx}}+\boldsymbol{{cotx}}+\boldsymbol{{secx}}\:\boldsymbol{{hence}} \\ $$$$\boldsymbol{{differentiate}}\:\boldsymbol{{the}}\:\boldsymbol{{function}} \\ $$
Answered by mindispower last updated on 12/Jan/21
$$\mathrm{1}={cos}^{\mathrm{2}} \left({x}\right)+{sin}^{\mathrm{2}} \left({x}\right) \\ $$$$\Leftrightarrow\frac{{cos}^{\mathrm{2}} \left({x}\right)+{sin}^{\mathrm{2}} \left({x}\right)+{sin}\left({x}\right)}{{cos}\left({x}\right){sin}\left({x}\right)} \\ $$$$=\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}+\frac{\left.{sinx}\right)}{{cos}\left({x}\right)}+\frac{\mathrm{1}}{{cos}\left({x}\right)} \\ $$$$={cot}\left({x}\right)+{tan}\left({x}\right)+{sec}\left({x}\right) \\ $$