Question

If $\left( \operatorname{secA}+\operatorname{tanA} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=\left( \operatorname{secA}-\operatorname{tanA} \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)$ prove that each is equal to $\pm 1$.

Answer 1

Hint: Relate the identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ with the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and express it as $\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)=1$. Put $\theta =A,B,C$ in this expression and multiply them and hence, use the given result in the problem to prove the given statement.

Complete step-by-step answer:
As we know the trigonometric identity related to $\sec ,\tan $ is given as
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ ………………… (i)
As, we know the algebraic identity of ${{a}^{2}}-{{b}^{2}}$ can be given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$………………… (ii)
So, applying the relation of equation (ii) with the equation (i), we get
$\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)=1$…………….(iii)
Now, coming to the question, it is given that the term $\left( \operatorname{secA}+\operatorname{tanA} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right),\left( \operatorname{secA}-\operatorname{tanA} \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)$ are equal to each other, then we need to prove that both will be equal to$\pm 1$.
So, we have
$\left( \operatorname{secA}+\operatorname{tanA} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=\left( \operatorname{secA}-\operatorname{tanA} \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)$ ………(iv)
Now, put $\theta =A,B,C$ in the equation (iii) to get the equations related to equation (iv).
So, we get equations as
If $\theta =A$
$\left( \sec A-\tan A \right)\left( \sec A+\operatorname{tanA} \right)=1$………… (v)
If $\theta =B$
$\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)=1$ ………………(vi)
If $\theta =C$
$\left( \operatorname{secC}-\operatorname{tanC} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=1$ ………………(vii)
Now, we can multiply the equations (v), (vi) and (vii) and hence, we get
$\left( \sec A-\tan A \right)\left( \sec A+\operatorname{tanB} \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=1$
Now, rearrange the terms just similar to equation (iv), hence, we can rewrite the above expression as
$\left[ \left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right) \right]\left[ \left( \sec A-\tan A \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right) \right]=1$ …………… (viii)
Now, we know that the values of terms written in big brackets of the above equations are equal from the equation (iv). So, we can replace one by another. So, we get
$\begin{align}
  & \Rightarrow \left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)\left( \sec A+\operatorname{tanA} \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \sec C+\operatorname{tanC} \right)=1 \\
 & \Rightarrow {{\left( \sec A+\tan A \right)}^{2}}{{\left( \operatorname{secB}+\operatorname{tanB} \right)}^{2}}{{\left( \operatorname{secC}+\operatorname{tanC} \right)}^{2}}=1 \\
 & \Rightarrow {{\left[ \left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right) \right]}^{2}}=1 \\
\end{align}$
And hence, taking square root to both sides, we get
$\left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=\pm 1$ …………….. (ix)
Now, we know that the above term in LHS will be equal to $\left( \sec A-\tan A \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)$ from the equation (iv). Hence, we get the values of both as
$\left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)=\left( \sec A-\tan A \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)=\pm 1$
Hence, it is proved.

Note: Another approach for solving the question would be that we can multiply both sides by
$\left( \sec A-\tan A \right)\left( \operatorname{secB}-\operatorname{tanB} \right)\left( \operatorname{secC}-\operatorname{tanC} \right)\Rightarrow \left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)$, And hence, we get
Let us multiply given relation by
$\left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right)$
Hence, we get
${{\left[ \left( \sec A+\tan A \right)\left( \operatorname{secB}+\operatorname{tanB} \right)\left( \operatorname{secC}+\operatorname{tanC} \right) \right]}^{2}}=\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)\left( {{\sec }^{2}}B-{{\tan }^{2}}B \right)\left( {{\sec }^{2}}C-{{\tan }^{2}}C \right)$
Where, use $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
Now, replace R.H.S of above equation by 1, with using trigonometric identity
$\left( {{\sec }^{2}}\theta +{{\tan }^{2}}\theta \right)=1$.
One may change $\sec \theta $ to $\dfrac{1}{\cos \theta },tan\theta $ to $\dfrac{\sin \theta }{cos\theta }$ and hence, can use identity ${{\sin }^{2}}\theta +co{{s}^{2}}\theta =1$ . So, it can be another approach for the question.
Observing the identity ${{\sin }^{2}}\theta -co{{s}^{2}}\theta =1$ of the form of $\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)=1$ is the key point of the question.