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

Hint: In this question we have to find the value of
$(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma )$ .
For this we will use the trigonometric identity to solve this question i.e.
${\sec ^2}\theta - {\tan ^2}\theta = 1$ .
So we will change the value of
$\theta = \alpha ,\beta ,\gamma $ one by one, and then we will find the value.

Complete step by step solution:
Here we have been given that
$(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma ) = \tan \alpha \tan \beta \tan \gamma $ .
We know the identity
${\sec ^2}\theta - {\tan ^2}\theta = 1$ .
First let us put the value of
$\theta = \alpha $
So we can write the identity as
${\sec ^2}\alpha - {\tan ^2}\alpha = 1$ .
By breaking the above expression we have
$(\sec \alpha + \tan \alpha )(\sec \alpha - \tan \alpha ) = 1$
We can write this also as
$\sec \alpha + \tan \alpha = \dfrac{1}{{\sec \alpha - \tan \alpha }}$
Similarly we will put the value of
 $\theta = \beta $
Again we will use the identity and by putting this value we have
 ${\sec ^2}\beta - {\tan ^2}\beta = 1$ .
By breaking the above expression we have
 $(\sec \beta + \tan \beta )(\sec \beta - \tan \beta ) = 1$
We can write this also as
$\sec \beta + \tan \beta = \dfrac{1}{{\sec \beta - \tan \beta }}$
Now let us put the value of
$\theta = \gamma $
So we can write the identity as
${\sec ^2}\gamma - {\tan ^2}\gamma = 1$ .
By breaking the above expression we have
$(\sec \gamma + \tan \gamma )(\sec \gamma - \tan \gamma ) = 1$
We can write this also as
$\sec \gamma + \tan \gamma = \dfrac{1}{{\sec \gamma - \tan \gamma }}$
Now by putting all these values in the given question, we can write
\[\dfrac{1}{{(\sec \alpha - \tan \alpha )}}\dfrac{1}{{(\sec \beta - \tan \beta )}}\dfrac{1}{{(\sec \gamma - \tan \gamma )}} = \tan \alpha \tan \beta \tan \gamma \] .
By cross multiplying the terms, we can write it as \[\dfrac{1}{{\tan \alpha }}\dfrac{1}{{\tan \beta }}\dfrac{1}{{\tan \gamma }} = (\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma )\] .
The above terms can also be written as
\[\cot \alpha \cot \beta \cot \gamma = (\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma )\] .
Hence the correct option is (a) $\cot \alpha \cot \beta \cot \gamma $.

Note:
We should note that in the above question we have used the algebraic identity to break the trigonometric identity into two terms i.e.
 ${a^2} - {b^2} = (a + b)(a - b)$ .
We should keep in mind that
$\dfrac{1}{{\tan \theta }} = \cot \theta $ . So we can write,
$\dfrac{1}{{\tan \alpha }} = \cot \alpha $ .
Similarly we can write
$\dfrac{1}{{\tan \beta }} = \cot \beta $ and,
$\dfrac{1}{{\tan \gamma }} = \cot \gamma $