Prove that Tan A÷1+sec A - tan A ÷1-sec A =2cosecA

1.	Prove that Tan A÷1+sec A - tan A ÷1-sec A =2cosecA
Answer» LHS =\xa0{tex}\\frac{\\tan A}{\\sec A-1}+\\frac{\\tan A}{\\sec A+1}{/tex}{tex}=\\frac{\\tan A(\\sec A+1)+\\tan A(\\sec A-1)}{(\\sec A-1)(\\sec A+1)}{/tex}{tex}=\\frac{\\tan A \\cdot \\sec A+\\tan A+\\tan A \\sec A-\\tan A}{\\sec ^{2} A-1}{/tex}{tex}=\\frac{2 \\tan A \\sec A}{\\tan ^{2} A}{/tex}\xa0[{tex}\\because{/tex}\xa0{tex}(sec^2\\theta - 1) = tan^2\\theta\xa0{/tex}]{tex}=\\frac{2 \\sec A}{\\tan A}{/tex}{tex}=\\frac{2 \\frac{1}{\\cos A}}{\\frac{\\sin A}{\\cos A}}{/tex}{tex}=2 \\times \\frac{1}{\\cos A} \\times \\frac{\\cos A}{\\sin A}{/tex}{tex}=\\frac{2}{\\sin \\mathrm{A}}{/tex}{tex}= 2 cosec\\ A{/tex} = RHS

Discussion