tan(x)-sin(x)=0.25 can be written sin(x)(1/cos(x)-1)=0.25. This is the same as 4sin(x)(1-cos(x))=cos(x). So 4sin(x)=cos(x)/(1-cos(x)). Square both sides: 16sin^2(x)=cos^2(x)/(1-cos(x))^2. That is: 16(1-cos^2(x))=cos^2(x)/(1-cos(x))^2. Let c=cos(x), then 16(1-c^2)=c^2/(1-c)^2 or 16(1-c^2)(1-c)^2=c^2 or 16(1+c)(1-c)^3=c^2.
When expanded this becomes: 16c^4-32c^3+c^2+32c-16=0. The solution must be between -1 and 1 because of the limitations of the cosine function. Let f(c)=16c^4-32c^3+c^2+32c-16. We can see that x=45 degrees is an approximate solution to tan(x)-sin(x)=0.25 because tan45=1 and sin45=cos45=sqrt(2)/2=0.71 (approx). The difference between tan45 and sin45 is 0.29 and we're looking for a difference of 0.25, so it must be close. Therefore the solution for c must be c close to 0.71. Using a calculator we can compute f(c) for values near to 0.71. When we do this we find that the solution to the quartic is approximately 0.73166, making x=cos^-1(0.73166)=42.974 degrees. When we substitute this angle into the original equation we get the result 0.2500006 as the difference between tangent and sine.