Question

Prove that each of the side is equal to $\pm 1$, for the given equation, $(\sec \text{A}+\tan \text{A})(\sec \text{B}+\tan \text{B})(\sec \text{C}+\tan \text{C})=(\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C})$.

Answer 1

To solve this question, one must remember the basic trigonometric identities. First of all we will multiply the whole equation with a trigonometric variable such that it becomes a trigonometric identity. After further solving it correctly, we will get the answer we expected or which we ned to prove in this question. The identity which we are going to use in this question is,
${{\sec }^{2}}\text{A}-{{\tan }^{2}}\text{A}=1$

Complete step by step answer:
The equation which is given in the question is,
$(\sec \text{A}+\tan \text{A})(\sec \text{B}+\tan \text{B})(\sec \text{C}+\tan \text{C})=(\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C})$
If we multiply with $(\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C})$on both the sides, then the equation will not be affected.
Now, we will see the result of the above equation which we will get after the multiplication,
The result is,
$\Rightarrow \left[ (\sec \text{A}+\tan \text{A})(\sec \text{B}+\tan \text{B})(\sec \text{C}+\tan \text{C}) \right]\left[ (\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C}) \right]$
$=\left[ (\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C}) \right]\left[ (\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C}) \right]$
$\Rightarrow \left[ ({{\sec }^{2}}\text{A}-{{\tan }^{2}}\text{A})({{\sec }^{2}}\text{B}-{{\tan }^{2}}\text{B})({{\sec }^{2}}\text{C}-{{\tan }^{2}}\text{C}) \right]={{\left[ (\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C}) \right]}^{2}}$
Here, the Left-Hand Side of the equation looks like an identity stated as follows,
$\Rightarrow {{\sec }^{2}}\text{A}-{{\tan }^{2}}\text{A}=1$
Now using the identity in the equation, we got above, the resultant of the equation will be,
$\Rightarrow 1={{\left[ (\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C}) \right]}^{2}}$
Now, taking the square root on both the sides, we will get,
$\Rightarrow \pm 1=(\sec \text{A}-\tan \text{A})(\sec \text{B}-\tan \text{B})(\sec \text{C}-\tan \text{C})$
Now comparing the final equation with the initial equation, we proved that was needed.

Note: The first thing we need to keep in mind while solving these types of questions, that the multiplication or division of any trigonometric variable is not fixed for any type, it differs for every question. So, solve the question carefully as a single mistake will make the complete answer incorrect, and it is very difficult to find the mistake after solving wrong, so check the solution with the solving steps, so that there are less chances of errors.