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

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,
$\dfrac{1}{{\tan \gamma }} = \cot \gamma$