Question

$Ifa{\sin }^2\alpha +b{\cos }^2\alpha =p,b{\sin }^2\beta +a{\cos }^2\beta =q,a\tan \alpha =b\tan \beta$. Show that $\frac{1}{a}+\frac{1}{b}=\frac{1}{p}+\frac{1}{q}wherea\ne p$ & all of them are nonzero.

Answer 1

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

$a{\tan }^2\alpha +b=p{\sec }^2\alpha \left(Dividingby{\cos }^2\alpha \right)$

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

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

${\tan }^2\alpha =\frac{p-b}{a-p}-\left(1\right)$

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

$=b{\tan }^2\beta +a=q{\sec }^2\beta \left(Dividingby{\cos }^2\beta \right)$

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

$=\left(b-q\right){\tan }^2\beta =q-a$

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

Dividing (1) by (2), we get

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

${\left(\frac{\tan \alpha }{\tan \beta }\right)}^2=\frac{pb-pq-b^2+bq}{aq-a^2-pq+pa}$

${\left(\frac{b}{a}\right)}^2=\frac{pb-pq-b^2+bq}{aq-a^2-pq+pa}$

$=b^{2}aq-b^2{a}^2-b^{2}pq+b^{2}pa=a^{2}pb-a^{2}pq-a^2{b}^2+a^{2}bq$

$=pq\left(a^2-b^2\right)+bpa\left(a-b\right)-baq\left(a-b\right)=0$

$=\left(a-b\right)\left[pq\left(a+b\right)-bpa-baq\right]=0$

$=pq\left(a+b\right)-bpa-baq=0\left(\because a\ne b\right)$

$=pqa+pqb=bqa+bqa$

$=\frac{1}{b}+\frac{1}{a}=\frac{1}{p}+\frac{1}{q}\left(dividebyabpq\right)$