Question 1117096
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1.  tan(2A) = {{{(2*tan(A))/(1-tan^2(A))}}} = {{{(2*(1/5))/(1-(1/5)^2)}}} = {{{((2/5))/(1-1/25)}}} = {{{((2/5))/(24/25))}}} = {{{(2*25)/(5*24)}}} = {{{5/12}}}.


    So, your calculations for tan(2A) are correct.




2.  To calculate  tan(4A),  notice that  4A = 2*(2A);  thus  tan(4A) = tan(2*(2A)).


    So, you can repeat THE SAME PROCEDURE  to get  tan(4A)  starting from  tan(2A), which you just know.




3.  tan(45°-4A) = {{{(tan(45^o)-tan(4A))/(1+tan(45^o)*tan(4A))}}} = {{{(1-tan(4A))/(1+tan(4A))}}}.


    So, when you will find  tan(4A)  (as I described in n.2), you will be in position to find  tan(45°-4A).
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