Question

If $(\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )(\sec \gamma +\tan \gamma )=\tan \alpha \tan \beta \tan \gamma$, then$(\sec \alpha -\tan \alpha )(\sec \beta -\tan \beta )(\sec \gamma -\tan \gamma )=$

• A. $cot\alpha cot\beta cot\gamma$
• B. $\tan \alpha \tan \beta \tan \gamma$
• C. $cot\alpha +cot\beta +cot\gamma$
• D. $\tan \alpha +\tan \beta +\tan \gamma$

Answer 1

The correct option is A

$cot\alpha cot\beta cot\gamma$

Explanation for the correct option

Step 1: Information required for the solution

Since $sec\alpha and\tan \alpha$are trigonometric identities

$se{c}^{2}x-{\tan }^{2}x=1$and $a^2-b^2=\left(a+b\right)\left(a-b\right)$

Step 2: Simplification and calculation of the given expression

Given expression as

$\left(sec\alpha +\tan \alpha \right)\left(sec\beta +\tan \beta \right)\left(sec\gamma +\tan \gamma \right)=\tan \alpha \tan \beta \tan \gamma .....\left(1\right)$,

Use the identity $se{c}^{2}x-{\tan }^{2}x=1$and $a^2-b^2=\left(a+b\right)\left(a-b\right)$

$$\begin{gathered} \left(sec\alpha +\tan \alpha \right)\left(sec\alpha -\tan \alpha \right)=1 \\ \left(sec\alpha -\tan \alpha \right)=\frac{1}{\left(sec\alpha +\tan \alpha \right)}.......\left(2\right) \\ \left(sec\beta +\tan \beta \right)\left(sec\beta -\tan \beta \right)=1 \\ \left(sec\beta -\tan \beta \right)=\frac{1}{\left(sec\beta +\tan \beta \right)}........\left(3\right) \\ \left(sec\gamma +\tan \gamma \right)\left(sec\gamma -\tan \gamma \right)=1 \\ \left(sec\gamma -\tan \gamma \right)=\frac{1}{\left(sec\gamma +\tan \gamma \right)}..........\left(4\right) \end{gathered}$$

Multiply $\left(2\right),\left(3\right)and\left(4\right)$ we get

$\begin{array}{rcl}\therefore \left(sec\alpha -\tan \alpha \right)\left(sec\beta -\tan \beta \right)\left(sec\gamma -\tan \gamma \right)& =& \left[\frac{1}{\left(sec\alpha +\tan \alpha \right)}\right].\left[\frac{1}{\left(sec\beta -\tan \beta \right)}\right].\left[\frac{1}{\left(sec\gamma -\tan \gamma \right)}\right]\\ & & \\ & =& \left[\frac{1}{\left(sec\alpha +\tan \alpha \right)\left(sec\beta +\tan \beta \right)\left(sec\gamma +\tan \gamma \right)}\right]\end{array}$

From equation $\left(1\right)$, we have

$$\begin{gathered} \Rightarrow \left(sec\alpha -\tan \alpha \right)\left(sec\beta -\tan \beta \right)\left(sec\gamma -\tan \gamma \right)=\frac{1}{\left(\tan \alpha \tan \beta \tan \gamma \right)} \\ \Rightarrow \left(sec\alpha -\tan \alpha \right)\left(sec\beta -\tan \beta \right)\left(sec\gamma -\tan \gamma \right)=cot\alpha cot\beta cot\gamma \end{gathered}$$

Hence, the correct option is $\left(A\right)$.