Question

If $tan\theta tan\varphi =\sqrt{(a-b)/(a+b)},$ prove that (a - b cos $\theta$ (a - b cos 2$\varphi$) is independent of $\theta$ and $\varphi$.

Answer 1

Let us put $tan\theta =t_1,tan\varphi =t_2$
$\therefore t_1^2{t}_2^2=\left(\frac{a-b}{a+b}\right)$
or $(a+b)t_1^2{t}_2^2=a-b$ ....(1)
Also $cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-t_1^2}{1+t_1^2}$ etc.
Now a - b cos $2\theta =a-b\frac{(1-t_1^2)}{(1+t_1^2)}=\frac{(a-b)+(a+b)t_1^2}{(1+t_1^2)}$
Put for (a - b) from (1),
$=\frac{a+b}{(1+t_1^2)}[t_1^2{t}_2^2+t_1^2]=\frac{(a+b)t_1^2(1+t_2^2)}{(1+t_1^2)},$
Similarly, a - b cos $2\varphi =\frac{(a+b)}{1+t_2^2}{t}_2^2(1+t_1^2)$
$\therefore$ (a - b cos $2\theta$) (a - b cos $2\varphi$) = $(a+b{)}^2$ $t_1^2{t}_2^2$
$=(a+b{)}^2\left\{\right(a-b\left)/\right(a+b\left)\right\}=a^2-b^2,$
which is independent of $\theta$.