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

Jul 19, 2016

See the proof given below.

#### Explanation:

Let $\sec \alpha = a , \sec \beta = b \mathmr{and} \sec \gamma = c$.

Since, $\forall \theta \in \mathbb{R} - \left\{\left(2 k + 1\right) \frac{\pi}{2} | k \in \mathbb{Z}\right\} , | \sec \theta | \ge 1$, we find

that, $| a | \ge 1 , | b | \ge 1 , | c | \ge 1$.

$\Rightarrow \tan \alpha = \pm \sqrt{{a}^{2} - 1} , \tan \beta = \pm \sqrt{{b}^{2} - 1}$, and,

$\tan \gamma = \pm \sqrt{{c}^{2} - 1}$ are defined as $| a | , | b | , | c | \ge 1$.

Then, by what is given, we have,

$a = b c \pm \sqrt{\left({b}^{2} - 1\right) \left({c}^{2} - 1\right)} \Rightarrow {\left(a - b c\right)}^{2} = \left({b}^{2} - 1\right) \left({c}^{2} - 1\right)$

$\Rightarrow {a}^{2} - 2 a b c + \cancel{{b}^{2} {c}^{2}} = \cancel{{b}^{2} {c}^{2}} - {b}^{2} - {c}^{2} + 1$

$\Rightarrow {b}^{2} - 2 a b c = 1 - {a}^{2} - {c}^{2}$

Adding ${a}^{2} {c}^{2}$ on both sides, we have,

${b}^{2} - 2 a b c + {a}^{2} {c}^{2} = {a}^{2} {c}^{2} - {a}^{2} - {c}^{2} + 1$

$\Rightarrow {\left(b - a c\right)}^{2} = {a}^{2} \left({c}^{2} - 1\right) - 1 \left({c}^{2} - 1\right) = \left({a}^{2} - 1\right) \left({c}^{2} - 1\right)$

$\Rightarrow b - a c = \pm \sqrt{\left({a}^{2} - 1\right) \left({c}^{2} - 1\right)}$

rArr b=ac+-sqrt((a^2-1)(c^2-1)

$\Rightarrow \sec \beta = \sec \alpha \sec \gamma \pm \tan \alpha \tan \gamma$.

Hence, the Proof. Enjoy Maths.!

Jul 19, 2016

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 "

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

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

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

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

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

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

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

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