Suggestion:

Since we would like to show:

If tan(45+x)+tan(45-x)=3 then show tan²(45+x)+tan²(45-x)=7

Let us try to square both sides of the starting equation:

tan(45+x)+tan(45-x)=3

Then we have:

tan^{2}(45+x)+ 2 tan(45+x)tan(45-x) + tan^{2}(45-x)=3^{2}

Two trig formulas which will be handy are:

tan(u + v) = (tan u + tan v) / (1 - tan u tan v)

tan(u - v) = (tan u - tan v) / (1 + tan u tan v)

Now let's look at the middle term:

2 tan(45+x)tan(45-x)

Also recall tan 45 = 1

By simplifying using the two trig formulas above, you will find that:

2 tan(45+x)tan(45-x) = 2

Then going back to the starting point we have:

tan^{2}(45+x)+ 2 tan(45+x)tan(45-x) + tan^{2}(45-x)=3^{2}

tan^{2}(45+x)+ 2 + tan^{2}(45-x)=9

Subtracting 2 from both sides we end up with we needed:

tan^{2}(45+x)+ tan^{2}(45-x)=7

Thanks,

Larry