Question

Let $\alpha=2 \tan ^{-1} \frac{1}{2}+\sin ^{-1} \frac{3}{5}$ and $\beta=\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\cot ^{-1} \frac{16}{63}$ be such that $2 \sin \alpha$ and $\cos \beta$ are roots of the equation $x^2-p x+q=0$, then find $(p-q)$.

Answer 1

$ \alpha=\frac{\pi}{2}, \beta=\pi $
$\therefore 2 \sin \alpha=2 \text { and } \cos \beta=-1$
$\text { So, p }=1, q=-2$