Question

Given $3\operatorname{Tan}A.\operatorname{Tan}B=1$ , show that $2\operatorname{Cos}\left( A+B \right)=\operatorname{Cos}\left( A-B \right)$?

Answer 1

Hint: From the question given that $3\operatorname{Tan}A.\operatorname{Tan}B=1$ from this we have to show that or we have to prove that $2\operatorname{Cos}\left( A+B \right)=\operatorname{Cos}\left( A-B \right)$. To solve the above question first we have to write the $\operatorname{Tan}A$as $\dfrac{\operatorname{Sin}A}{\operatorname{Cos}A}$ and then we have to shift to the right-hand side. There we have to apply the componendo and dividendo it means if the ratio of a to b is equal to the ratio of c to d, then the ratio of $a+b$ to $a-b$ is equal to the ratio of $c+d$ to $c-d$ .this property is called as componendo and dividendo. By applying this we will get the required answer.

Complete step-by-step answer:
From the given question we know that
$\Rightarrow 3\operatorname{Tan}A.\operatorname{Tan}B=1$
Now we will write the Tan in terms of Sin and Cos, as we know that
 $\Rightarrow \operatorname{Tan}A=\dfrac{\operatorname{Sin}A}{\operatorname{Cos}A}$
Therefore, by this we can write the above equation as,
$\Rightarrow 3\dfrac{\operatorname{Sin}A\times \operatorname{Sin}B}{\operatorname{Cos}A\times \operatorname{Cos}B}=1$
Now we have to shift that trigonometric functions from left hand side to the right hand side,
By shifting we will get,
$\Rightarrow 3=\dfrac{\operatorname{Cos}A\times \operatorname{Cos}B}{\operatorname{Sin}A\times \operatorname{Sin}B}$
Here we have to apply the componendo and dividendo as we know that if the ratio of a to b is equal to the ratio of c to d, then the ratio of $a+b$ to $a-b$ is equal to the ratio of $c+d$ to $c-d$ . that means,
$\Rightarrow \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$
this property is called as componendo and dividendo.
By applying componendo and dividendo we will get,
$\Rightarrow \dfrac{3+1}{3-1}=\dfrac{\operatorname{Cos}A\times \operatorname{Cos}B+\operatorname{Sin}A\times \operatorname{Sin}B}{\operatorname{Cos}A\times \operatorname{Cos}B-\operatorname{Sin}A\times \operatorname{Sin}B}$
By further simplifying we will get,
$\Rightarrow \dfrac{4}{2}=\dfrac{\operatorname{Cos}\left( A-B \right)}{\operatorname{Cos}\left( A+B \right)}$
 By further simplifying we will get,
$\Rightarrow 2\operatorname{Cos}\left( A+B \right)=\operatorname{Cos}\left( A-B \right)$
Therefore, hence proved.

Note: Students should recall all the formulas of trigonometry while doing the above problem, students should know how to and when the concept or property of componendo and dividendo should be applied.