If sec A - tan A=4, THEN PROVE THAT cos A= 8/16

1.	If sec A - tan A=4, THEN PROVE THAT cos A= 8/16
Answer» Now, sec A - tan A = 4........(i)We know that(sec A - tan A)(sec A + tan A) = 1{tex}\\Rightarrow{/tex}\xa04(secA + tanA) = 1{tex}\\Rightarrow{/tex}\xa0secA + tanA = {tex}\\frac{1}{4}{/tex}\xa0.......(ii)Adding\xa0(i) and (ii) we get,2 secA =\xa0{tex}\\frac{{17}}{4}{/tex}{tex}\\Rightarrow{/tex}\xa0secA =\xa0{tex}\\frac{{17}}{8}{/tex}{tex}\\Rightarrow{/tex}\xa0cosA = {tex}\\frac{8}{{17}}{/tex}

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