Question

√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1

Answer 1

`Use:\\\\\sec A=(1)/(\cos A)\\\\\tan A=(\sin A)/(\cos A)\\\\\csc A=(1)/(\sin A)\\\\√(ab)=√(a)\cdot√(b)\\\\\sqrt{(a)/(b)}=(√(a))/(√(b))\\---------------------------------\\\\\sec A+\tan A=(1)/(\cos A)+(\sin A)/(\cos A)=(1+\sin A)/(\cos A)\\\\\sec A-\tan A=(1-\sin A)/(\cos A)\\\\\csc A-1=(1)/(\sin A)-(\sin A)/(\sin A)=(1-\sin A)/(\sin A)\\\\\csc A+1=(1+\sin A)/(\sin A)`

`(√(\sec A+\tan A))/(√(\sec A-\tan A))\cdot(√(\cos A-1))/(√(\cos A+1))=1\\\\L_s=\sqrt{(\sec A+\tan A)/(\sec A-\tan A)\cdot(\cos A-1)/(\cos A+1)}=\sqrt{((1+\sin A)/(\cos A))/((1-\sin A)/(\cos A))\cdot((1-\sin A)/(\sin A))/((1+\sin A)/(\sin A))}\\\\=\sqrt{(1+\sin A)/(\cos A)\cdot(\cos A)/(1-\sin A)\cdot(1-\sin A)/(\sin A)\cdot(\sin A)/(1+\sin A)}\\\\\text{Everything are simplified}\\\\=√(1)=1=R_s`