Question 1151777
<pre>
{{{sec(A) + tan(A) = 4}}}

{{{1/cos(A) + sin(A)/cos(A) = 4}}}

{{{1+sin(A)=4cos(A)}}}

{{{1+sin(A)=4cos(A)}}}

{{{sin(A)=4cos(A)-1}}}

{{{sin^2(A)=(4cos(A)-1)^2}}}

{{{sin^2(A)=16cos^2(A)-8cos(A)+1}}}

{{{1-cos^2(A)=16cos^2(A)-8cos(A)+1}}}

{{{8cos(A)=17cos^2(A)}}}


{{{8cos(A)-17cos^2(A)=0}}}

{{{cos(A)(8-17cos(A))=0}}}

cos(A)=0;  8-17cos(A)=0
   A=90°;   -17cos(A)=-8
               {{{cos(A)=(-8)/(-17)}}}
               {{{cos(A)=8/17}}}

We discard 90° because neither sec(90°) nor tan(90°) 
which occur in the original equation are defined.

Answer: {{{8/17}}}

Edwin</pre>