Question

If $\frac{p\cos (x+\alpha )+q\sin (x+\beta )}{p^{\prime }\sin (x+\alpha )+q^{\prime }\cos (x+\beta )}$ has the same value at $x=0$ and $x=\pi /2,$ then $p{p}^{\prime }-q{q}^{\prime }-(p^{\prime }q-p{q}^{\prime })\sin (\alpha -\beta )=$

Answer 1

$\frac{p\cos \left(\alpha \right)+q\sin \beta }{p^{\prime }\sin \alpha +q^{\prime }\cos \beta }=\frac{-\left(p\sin \alpha \right)+\left(q\cos \beta \right)}{p^{\prime }\cos \alpha -q^{\prime }\sin \beta }$

$\Rightarrow p{p}^{\prime }{\cos }^2\alpha -p{q}^{\prime }\cos \alpha \sin \beta +p^{\prime }q\cos \alpha \sin \beta -q{q}^{\prime }{\sin }^2\beta$

$=-p{p}^{\prime }{\sin }^2\alpha +p^{\prime }q\sin \alpha \cos \beta -p{q}^{\prime }\sin \alpha \cos +q{q}^{\prime }{\cos }^2\beta$

$\Rightarrow p{p}^{\prime }-q{q}^{\prime }+p{q}^{\prime }\left(\sin \right(\alpha -\beta \left)\right)+p^{\prime }q\left(\sin \right(\beta -\alpha \left)\right)=0$