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 $a\ne p$ and all of them are nonzero.

Answer 1

Given $a{\sin }^2\alpha +b{\cos }^2\alpha =p$....equation 1

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

$a^2{\tan }^2\alpha =b^2{\tan }^2\beta$.....equation 3

Divide equation 1 by ${\cos }^2\alpha$ :

$a^2{\tan }^2\alpha +b=p{\sec }^2\alpha$

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

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

Dividing equation 2 by ${\cos }^2\beta$ :

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

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

Putting values of ${\tan }^2\alpha$ and ${\tan }^2\beta$ in equation 3:

$a^2\left(\frac{p-b}{a-p}\right)=b^2\left(\frac{q-a}{b-q}\right)$

$a^2\left[(p-b)(b-q)\right]=b^2\left[(q-a)(a-p)\right]$

$a^{2}bp-pq{a}^2+a^{2}bq-a^2{b}^2=b^{2}aq-b^{2}pq+b^{2}ap-b^2{a}^2$

$abp(a-p)+abq(a-b)=pq(a^2-b^2)$

$abp+abq=pq(a+b)$

$(p+q)ab=pq(a+b)$

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