Question 985833
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Find the exact value of {{{tan(arccos(-3/5))}}}.
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Let  {{{alpha}}} = {{{arccos(-3/5)}}}.     (Notice that  {{{alpha}}}  lies in the segment  [{{{0}}}, {{{pi}}}]).   We need to find  {{{tan(alpha)}}}. 


Then  {{{cos(alpha)}}} = {{{-3/5}}},


{{{sin(alpha)}}} = {{{sqrt(1 - cos^2(alpha))}}} = {{{sqrt(1 - (-3/5)^2)}}} = {{{sqrt(1 - (9/25))}}} = {{{sqrt((25-9)/25)}}} = {{{sqrt(16/25)}}} = {{{4/5}}}. 


Now   {{{tan(alpha)}}} = {{{sin(alpha)/cos(alpha)}}} = {{{4/5}}}:{{{(-3/5)}}} = {{{-4/3}}}.