Question Number 106118 by DeepakMahato last updated on 02/Aug/20
Commented by DeepakMahato last updated on 02/Aug/20
$${Please}\:{Solve}... \\ $$
Answered by nimnim last updated on 02/Aug/20
$${A}+{B}=\mathrm{90}\Rightarrow{A}=\mathrm{90}−{B}\:\:{or}\:{B}=\mathrm{90}−{A} \\ $$$${LHS}:\:\sqrt{\frac{{tanAtanB}+{tanAcotB}}{{sinAsecB}}−\frac{{sin}^{\mathrm{2}} {B}}{{cos}^{\mathrm{2}} {A}}} \\ $$$$\:\:=\sqrt{\frac{{tanAtan}\left(\mathrm{90}−{A}\right)+{tanAcot}\left(\mathrm{90}−{A}\right)}{{sinAsec}\left(\mathrm{90}−{A}\right)}−\frac{{sin}^{\mathrm{2}} \left(\mathrm{90}−{A}\right)}{{cos}^{\mathrm{2}} {A}}} \\ $$$$\:\:=\:\sqrt{\frac{{tanA}.{cotA}+{tanAtanA}}{{sinA}.{cosecA}}−\frac{{cos}^{\mathrm{2}} {A}}{{cos}^{\mathrm{2}} {A}}} \\ $$$$\:\:\:=\:\sqrt{\frac{\mathrm{1}+{tan}^{\mathrm{2}} {A}}{\mathrm{1}}−\mathrm{1}}\:=\:\sqrt{{sec}^{\mathrm{2}} {A}−\mathrm{1}} \\ $$$$\:\:\:=\:\sqrt{{tan}^{\mathrm{2}} {A}}\:\:=\:\:{tanA}\:\left({RHS}\right) \\ $$
Commented by DeepakMahato last updated on 02/Aug/20
Thank You Once Again...