# Given secalpha=secbetasecgamma+tanbetatangamma How will you show? secbeta =secgammasecalpha+-tangammatanalpha

Jul 2, 2016

As below

#### Explanation:

Given relation

$\sec \alpha = \sec \beta \sec \gamma + \tan \beta \tan \gamma$

$\implies \sec \alpha - \sec \beta \sec \gamma = \tan \beta \tan \gamma$

$\textcolor{g r e e n}{\text{Squaring both sides}}$

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

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

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

$\implies - 2 \sec \alpha \sec \beta \sec \gamma = - {\sec}^{2} \alpha + \left({\sec}^{2} \beta - 1\right) \left({\sec}^{2} \gamma - 1\right) - {\sec}^{2} \beta {\sec}^{2} \gamma$

$\implies - 2 \sec \alpha \sec \beta \sec \gamma = - {\sec}^{2} \alpha + \cancel{{\sec}^{2} \beta {\sec}^{2} \gamma} + 1 - {\sec}^{2} \gamma - {\sec}^{2} \beta - \cancel{{\sec}^{2} \beta {\sec}^{2} \gamma}$

$\implies {\sec}^{2} \beta - 2 \sec \alpha \sec \beta \sec \gamma = - {\sec}^{2} \alpha + 1 - {\sec}^{2} \gamma$

color(blue)("Adding "(sec^2gammasec^2alpha)" both sides "

$\implies {\sec}^{2} \beta - 2 \sec \alpha \sec \beta \sec \gamma + {\sec}^{2} \gamma {\sec}^{2} \alpha = {\sec}^{2} \gamma {\sec}^{2} \alpha - {\sec}^{2} \alpha + 1 - {\sec}^{2} \gamma$

$\implies {\left(\sec \beta - \sec \gamma \sec \alpha\right)}^{2} = {\sec}^{2} \alpha \left({\sec}^{2} \gamma - 1\right) - 1 \left({\sec}^{2} \gamma - 1\right)$

$\implies {\left(\sec \beta - \sec \gamma \sec \alpha\right)}^{2} = \left({\sec}^{2} \gamma - 1\right) \left({\sec}^{2} \alpha - 1\right)$

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

$\implies \left(\sec \beta - \sec \gamma \sec \alpha\right) = \pm \sqrt{{\tan}^{2} \gamma {\tan}^{2} \alpha}$

$\implies \sec \beta - \sec \gamma \sec \alpha = \pm \tan \gamma \tan \alpha$

$\implies \textcolor{B L U E}{\sec \beta = \sec \gamma \sec \alpha \pm \tan \gamma \tan \alpha}$

Shown