Question Number 227900 by needothink last updated on 26/Feb/26
$$\boldsymbol{\mathrm{Q}}.\:\frac{\mathrm{1}}{\mathrm{sec}\:\boldsymbol{\mathrm{x}}−\mathrm{tan}\:\boldsymbol{\mathrm{x}}\:}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\boldsymbol{\mathrm{x}}}\:−\frac{\mathrm{1}}{\mathrm{sec}\:\boldsymbol{\mathrm{x}}+\mathrm{tan}\:\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{it}} \\ $$
Answered by som(math1967) last updated on 26/Feb/26
$$\frac{\mathrm{1}}{{secx}−{tanx}}+\frac{\mathrm{1}}{{secx}+{tanx}} \\ $$$$=\frac{{secx}+{tanx}+{secx}−{tanx}}{{sec}^{\mathrm{2}} {x}−{tan}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{2}{secx}}{\mathrm{1}}=\frac{\mathrm{2}}{{cosx}} \\ $$$$\therefore\frac{\mathrm{1}}{{secx}−{tanx}}+\frac{\mathrm{1}}{{secx}+{tanx}}=\frac{\mathrm{2}}{{cosx}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{secx}−{tanx}}\:−\frac{\mathrm{1}}{{cosx}}=\frac{\mathrm{1}}{{cosx}}−\frac{\mathrm{1}}{{secx}+{tanx}} \\ $$