Question

If $a{\sin }^2\alpha +b{\cos }^2\alpha =P$, $b{\sin }^2\beta +a{\cos }^2\beta =q$, $a\tan \alpha =b\tan \beta$ show that $\frac{1}{a}+\frac{1}{b}=\frac{1}{p}+\frac{1}{q}$ where $a\ne p$.

Answer 1

$a{\sin }^2\alpha +b{\cos }^2\alpha =p$

divide both sides by ${\cos }^2\alpha$

$a{\tan }^2\alpha +b=p{\text{sec}}^2\alpha =p(1+{\tan }^2\alpha )$

$(a-p){\tan }^2\alpha =(p-b)$

${\tan }^2\alpha =\frac{p-b}{a-p}$

$b{\sin }^2\beta +b{\cos }^2\beta =q$

divide both sides by ${\sin }^2\beta$

$b{\tan }^2\beta +a=q{\text{sec}}^2\beta =q(1+{\tan }^2\beta )$

${\tan }^2\beta =\frac{q-a}{b-q}$

$(a\tan \alpha {)}^2=(b\tan \beta {)}^2$

$\frac{{\tan }^2\alpha }{{\tan }^2\beta }=\frac{b^2}{a^2}$

from the above equations

$a^2\times [pb-b^2+qb-qp]=b^2\times [aq-pq-a^2+ap]$

on simplifying

$abp+aqb=pq\times (a+b)$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{p}+\frac{1}{q}$