Another question proving that a trig equation follows from another given trig equatio

Hello,

Square the first equation :
$a^2\cos^2\theta+b^2\sin^2\theta-2ab\cos\theta\sin\theta=c^2$
Thus $2ab\cos\theta\sin\theta={\color{red}a^2\cos^2\thet a+b^2\sin^2\theta-c^2}$

Then $(a\sin\theta+b\cos\theta)^2=a^2\sin^2\theta+b^2\co s^2\theta+2ab\cos\theta\sin\theta$
$=a^2\sin^2\theta+b^2\cos^2\theta+{\color{red}a^2\c os^2\theta+b^2\sin^2\theta-c^2}$
$=a^2(\cos^2\theta+\sin^2\theta)+b^2(\cos^2\theta+\ sin^2\theta)-c^2$

And this gives you the result, using the identity $\cos^2\theta+\sin^2\theta=1$