Question

If \[k = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\], then $k$ is equal to?
A) $0$
B) $1$
C) $ \pm 3$
D) $ \pm 4$

Answer 1

Hint: Given the value of $k$ is equivalent to the product of six terms in trigonometric ratios $\sec $ and $\tan $. We can change the order of these terms in the equation. Thus we can collect the terms of $A,B$ and $C$ separately. There we can apply the formula of $(a + b)(a - b)$. For simplifying we can use the relation between $\tan $ and $\sec $ of an angle. Thus we get the right answer.

Formula used:
For any angle $\theta $ we have,
${\sec ^2}\theta - {\tan ^2}\theta = 1$
For any $a,b$ we have,
$(a + b)(a - b) = {a^2} - {b^2}$

Complete step-by-step answer:
Given that, \[k = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\]
Since multiplication is commutative ( means $a \times b = b \times a$), we can rearrange the terms in the above equation.
Rearrange the terms in such a way that we can collect terms of $A,B$ and $C$ separately.
This gives,
\[k = (\sec A + \tan A)(\sec A - \tan A)(\sec B + \tan B)(\sec B - \tan B)(\sec C + \tan C)(\sec C - \tan C)\]
Now consider \[[(\sec A + \tan A)(\sec A - \tan A)][(\sec B + \tan B)(\sec B - \tan B)][(\sec C + \tan C)(\sec C - \tan C)]\]
Now we have, $(a + b)(a - b) = {a^2} - {b^2}$
So we can write,
\[(\sec A + \tan A)(\sec A - \tan A) = {\sec ^2}A - {\tan ^2}A\]
\[(\sec B + \tan B)(\sec B - \tan B) = {\sec ^2}B - {\tan ^2}B\]
\[(\sec C + \tan C)(\sec C - \tan C) = {\sec ^2}C - {\tan ^2}C\]
Therefore we get our equation as,
\[k = ({\sec ^2}A - {\tan ^2}A)({\sec ^2}B - {\tan ^2}B)({\sec ^2}C - {\tan ^2}C)\]
For any angle $\theta $ we have,
${\sec ^2}\theta - {\tan ^2}\theta = 1$
Applying this result in the above equation we have,
\[k = 1 \times 1 \times 1\]
$ \Rightarrow$ $k = 1$
$\therefore $ The answer is option B.

Additional information:
We have more similar relations in trigonometry:
For any angle $\theta $ we have,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${\text{cose}}{{\text{c}}^2}\theta - {\cot ^2}\theta = 1$


Note: This question may seem to be complicated in the first look. But we have to observe it carefully for rearranging and applying the results. Identifying the method is the important step here. A main thing to be noted is that ${\sec ^2}\theta - {\tan ^2}\theta = 1$ holds for the same $\theta $. That is, ${\sec ^2}x - {\tan ^2}y$ need not equal to one if $x \ne y$.
Also if instead of multiplication, we were given the sum of these terms, we could not have used the result.