Question Number 189375 by moh777 last updated on 15/Mar/23
$$ \\ $$$$\:\:\:\:{show}\:{that}\:: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{cscx}\:+\:{cot}\:{x}}\:=\:{cscx}\:−\:{cot}\:{x} \\ $$$$ \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 15/Mar/23
$$\frac{\mathrm{1}}{\mathrm{csc}{x}+\mathrm{cot}{x}}=\frac{\mathrm{sin}{x}}{\mathrm{1}+\mathrm{cos}{x}}=\frac{\mathrm{sin}{x}\left(\mathrm{1}−\mathrm{cos}{x}\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{sin}{x}\left(\mathrm{1}−\mathrm{cos}{x}\right)}{\mathrm{sin}^{\mathrm{2}} {x}}=\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{sin}{x}} \\ $$$$=\mathrm{csc}{x}−\mathrm{cot}{x} \\ $$
Answered by BaliramKumar last updated on 15/Mar/23
$${LHS}\:=\:\frac{\mathrm{1}}{{cosecx}\:+\:{cotx}}\:=\:\frac{{cosec}^{\mathrm{2}} {x}\:−\:{cot}^{\mathrm{2}} {x}}{{cosecx}\:+\:{cotx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\cancel{\left({cosecx}\:+\:{cotx}\right)}\left({cosecx}\:−\:{cotx}\right)}{\cancel{\left({cosecx}\:+\:{cotx}\right)}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{cosecx}\:−\:{cotx}\:=\:{RHS} \\ $$$$ \\ $$
Answered by manxsol last updated on 15/Mar/23
$${recall}\:\:\:\:\:{cosec}^{\mathrm{2}} =\mathrm{1}+{cot}^{\mathrm{2}} {x} \\ $$$$\:\frac{\mathrm{1}}{{cscx}\:+\:{cot}\:{x}}\:×\frac{{cscx}−{cotx}}{{cscx}−{cotx}}= \\ $$$$\frac{{cscx}−{cotx}}{{csc}^{\mathrm{2}} {x}−{ct}^{\mathrm{2}} {x}}={cscx}−{cotx} \\ $$