Question

Let $cos\alpha =cos\beta cos\varphi =cos\gamma cos\theta$,
$sin\alpha =2sin\left(\varphi /2\right)sin\left(\theta /2\right)$
Prove that $ta{n}^2\left(\alpha /2\right)=ta{n}^2\left(\beta /2\right)\cdot ta{n}^2\left(\gamma /2\right)$

Answer 1

We have to eliminate $\theta$ and $\varphi$
$si{n}^2\alpha =4si{n}^2\left(\varphi /2\right)\cdot si{n}^2\left(\varphi /2\right)=(1-cos\varphi )(1-cos\theta )$
or $1-co{s}^2\alpha =(1-\frac{cos\alpha }{cos\beta })(1-\frac{cos\alpha }{cos\beta })$
$=1-cos\alpha (\frac{1}{cos\beta }+\frac{1}{cos\gamma })+\frac{co{s}^2\alpha }{cos\beta cos\gamma }$
or $cos\alpha \left(\frac{cos\beta +cos\gamma }{cos\beta cos\gamma }\right)=co{s}^2\alpha \cdot \frac{1+cos\beta cos\gamma }{cos\beta cos\gamma }$
$\therefore \frac{cos\beta +cos\gamma }{cos\beta cos\gamma }=\frac{cos\alpha }{1}$
Apply compo. and divi.
$\frac{1-cos\beta -cos\gamma +cos\beta cos\gamma }{1+cos\beta -cos\gamma +cos\beta cos\gamma }=\frac{1-cos\alpha }{1+cos\alpha }$
or $\frac{(1-cos\beta )(1-cos\gamma )}{(1+cos\beta )(1+cos\gamma )}=\frac{1-cos\alpha }{1+cos\alpha }$
or $\frac{2si{n}^2\left(\beta /2\right)\cdot 2si{n}^2\left(\gamma /2\right)}{2co{s}^2\left(\beta /2\right)\cdot 2co{s}^2\left(\gamma /2\right)}=\frac{2si{n}^2\left(\alpha /2\right)}{2co{s}^2\left(\alpha /2\right)}$
or $ta{n}^2\left(\beta /2\right)ta{n}^2\left(\gamma /2\right)=ta{n}^2\left(\alpha /2\right)$