Question Number 185647 by Mingma last updated on 24/Jan/23
Answered by mahdipoor last updated on 24/Jan/23
$${secx}+{tan}^{\mathrm{3}} {x}.{cscx}=\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}=\frac{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}= \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{3}} {x}}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\mathrm{3}/\mathrm{2}} =\left(\mathrm{2}−{e}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$