False proof of Sin A + Sin B = 1 in a right angled triangle

Question

Assume a right angled triangle ABC where C is right angled. Small letters a,b and c should be opposite sides of the angles A B and C respectively.

This gives us

Sin A = $\frac{a}{c}$ ,Sin B = $\frac{b}{c}$

Since $a^2 + b^2 = c^2$

Sin A = $\frac{a^2}{a^2 + b^2}$ ,Sin B = $\frac{b^2}{a^2 + b^2}$

So, Sin A + Sin B

=$\frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2}$

=$\frac{a^2+b^2}{a^2 + b^2}$

=1

Obviously, for a 3-4-5 right angled triangle, Sin A + Sin B = $\frac {7}{5}$

So, what is wrong with this proof?

Answer 1

You don't have $Sin A = \frac{a^2}{a^2+b^2}$.

What you have is $(Sin A)^2 = \frac{a^2}{a^2+b^2}$. Thus you prove $Sin^2+Cos^2 = 1$, which is true.