Question

Tan A + sin A/tan A-sin A
= sec A +1/sec A-1

Answer 1

We need to prove

`(\tan A+\sin A)/(\tan A-\sin A)=(\sec A+1)/(\sec A-1)`

`\text{LHS}=(\tan A+\sin A)/(\tan A-\sin A)\\\\=((\sin A)/(\cos A)+\sin A)/((\sin A)/(\cos A)-\sin A)\\\\=(\sin A+\sin A \cos A)/(\sin A-\sin A \cos A)\\\\=(\sin A(1+\cos A))/(\sin A(1-\cos A))\\\\=(1+\cos A)/(1-\cos A)\\\\=(1+(1)/(\sec A))/(1-(1)/(\sec A))\\\\=(\sec A+1)/(\sec A-1)\\\\ =\text{RHS}`