Prove that (tan A + sin A)/(tan A - sin A) = (sec A + 1)/(sec A - 1).�

1.	Prove that (tan A + sin A)/(tan A - sin A) = (sec A + 1)/(sec A - 1).
Answer» $\huge\underline{\rm{☆To \ Prove}}$ $\rm{\dfrac{\ tanA + \ sinA}{\ tanA - \ sinA} = \dfrac{\ secA + 1}{\ secA - 1}}$ $\\$ $\rm{LHS = \dfrac{\ tanA + \ sinA}{\ tanA - \ sinA}} \\ \\$ $\rm{ = \dfrac{\dfrac{\ sinA}{\ cosA} + \ sinA}{\dfrac{\ sinA}{\ cosA} - \ sinA}} \\ \\$ $\rm{ = \dfrac{ \dfrac{ \ sinA + \ sinA \ cosA }{ \ cosA }}{\dfrac{ \ sinA - \ sinA \ cosA}{ \ cosA}}} \\ \\$ $\rm{ = \dfrac{\ sinA + \ sinA\ cosA}{\ sinA - \ sinA \ cosA}} \\ \\$ $\rm{ = \dfrac{\ sinA(1 + \ cosA}{\ sinA(1 - \ cosA}} \\ \\$ $\rm{ = \dfrac{\ sinA + \ dfrac{1}{\ secA}}{\ sinA - \dfrac{1}{\ secA}}} \\ \\$ $\rm{ = \dfrac{\dfrac{\ secA + 1}{\ secA}}{\dfrac{\ secA - 1}{\ secA}}} \\ \\$ $\rm{ = \dfrac{\ secA + 1}{\ secA - 1}= RHS} \\ \\$ $\red{\boxed{\rm{Hence \ Proved✔}}}$ $\\$ #NAWABZAADI