Question

Given that the angles $\alpha ,\beta ,\gamma$ are connected by the relation -
$2ta{n}^2\alpha ta{n}^2\beta ta{n}^2\gamma +ta{n}^2\alpha ta{n}^2\beta +ta{n}^2\beta ta{n}^2\gamma +ta{n}^2\gamma ta{n}^2\alpha =1$
find the value of $si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma$

Answer 1

We have 2 = 1 + 1
$ta{n}^2\beta ta{n}^2\gamma (1+ta{n}^2\alpha )+ta{n}^2\gamma ta{n}^2\alpha (1+ta{n}^2\beta )=1-ta{n}^2\alpha ta{n}^2\beta$
or $\frac{si{n}^2\beta si{n}^2\gamma +si{n}^2\gamma si{n}^2\alpha }{co{s}^2\alpha co{s}^2\beta co{s}^2\gamma }=\frac{co{s}^2\alpha co{s}^2\beta -si{n}^2\alpha si{n}^2\beta }{co{s}^2\alpha co{s}^2\beta }$
$si{n}^2\beta si{n}^2\gamma +si{n}^2\gamma si{n}^2\alpha =(1-si{n}^2\gamma ).$
$\left[\right(1-si{n}^2\alpha \left)\right(1-si{n}^2\beta )-si{n}^2\alpha si{n}^2\beta ]$
or $si{n}^2\beta si{n}^2\gamma +si{n}^2\gamma si{n}^2\alpha =(1-si{n}^2\gamma )[1-si{n}^2\alpha -si{n}^2\beta ]$
or $0=1-si{n}^2\gamma -si{n}^2\alpha -si{n}^2\beta$
or $\sum si{n}^2\alpha =1.$