Prove that tan A + sec a - 1 / 10 a minus sec a + 1 is equals to 1 + sin a divided by Cos A

Prove that tan A + sec a - 1 / 10 a minus sec a + 1 is equals to 1 + s

1.	Prove that tan A + sec a - 1 / 10 a minus sec a + 1 is equals to 1 + sin a divided by Cos A
Answer» {tex}\\frac{{\\tan A + \\sec A - 1}}{{\\tan A - \\sec A + 1}} = \\frac{{\\tan A + \\sec A - ({{\\sec }^2}A - {{\\tan }^2}A)}}{{\\tan A - \\sec A + 1}}{/tex}\xa0[{tex}\\because{/tex}\xa01 + tan2A = sec2A\xa0{tex}\\Rightarrow{/tex}\xa0sec2A - tan2A = 1]{tex} = \\frac{{\\tan A + \\sec A - \\{ (\\sec A - \\tan A)(\\sec A + \\tan A)\\} }}{{\\tan A - \\sec A + 1}}{/tex}Now, take (tanA + secA) as a common term, we get{tex} = \\frac{{(\\tan A + \\sec A)(1 - \\sec A + \\tan A)}}{{\\tan A - \\sec A + 1}}{/tex}= tanA + secA{tex} = \\frac{{\\sin A}}{{\\cos A}} + \\frac{1}{{\\cos A}}{/tex}{tex} = \\frac{{1 + \\sin A}}{{\\cos A}}{/tex}{tex}\\therefore{/tex} Hence proved