Question

If $\sec \alpha =\sec \beta \sec \gamma +\tan \beta \tan \gamma$ then show that $\sec \beta =\sec \gamma \sec \alpha \pm \tan \gamma \tan \alpha$

Answer 1

$\text{sec}\alpha =\text{sec}\beta \text{sec}\gamma +\tan \beta \tan \gamma ⟹\tan \beta \tan \gamma =\text{sec}\alpha -\text{sec}\beta \text{sec}\gamma$

squaring on both sides

${\tan }^2\beta {\tan }^2\gamma ={\text{sec}}^2\alpha +{\text{sec}}^2\beta {\text{sec}}^2\gamma -2\text{sec}\alpha \text{sec}\beta \text{sec}\gamma ⟹({\text{sec}}^2\beta -1){\tan }^2\gamma$

$={\text{sec}}^2\alpha +{\text{sec}}^2\beta {\text{sec}}^2\gamma -2\text{sec}\alpha \text{sec}\beta \text{sec}\gamma$

$({\text{sec}}^2\gamma -{\tan }^2\gamma ){\text{sec}}^2\beta -\left(2\text{sec}\alpha \text{sec}\gamma \right)\text{sec}\beta +({\text{sec}}^2\alpha +{\tan }^2\gamma )=0$

$⟹{\text{sec}}^2\beta -\left(2\text{sec}\alpha \text{sec}\gamma \right)\text{sec}\beta +({\text{sec}}^2\alpha +{\tan }^2\gamma )=0$

this is in the form of $a{x}^2+bx+c=0$ whose roots are$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$⟹\text{sec}\beta =\frac{2\text{sec}\alpha \text{sec}\gamma \pm \sqrt{4{\text{sec}}^2\alpha {\text{sec}}^2\gamma -4({\tan }^2\gamma +{\text{sec}}^2\alpha )}}{2\left(1\right)}=\text{sec}\alpha \text{sec}\gamma \pm \sqrt{{\text{sec}}^2\alpha ({\text{sec}}^2\gamma -1)-{\tan }^2\gamma }$

$=\text{sec}\alpha \text{sec}\gamma \pm \sqrt{{\tan }^2\gamma ({\text{sec}}^2\alpha -1)}=\text{sec}\alpha \text{sec}\gamma \pm \tan \alpha \tan \gamma$