Question

If \[\tan \left( A+B \right)=x\ and\ \tan \left( A-B \right)=y\], find the values of tan2A and tan2B.

Answer 1

Hint: We will be using the concept of trigonometric identities to solve the problem. We will write tan2A as $\tan \left( \left( A+B \right)+\left( A-B \right) \right)$ then will apply the formula for $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ and substitute the value of \[\tan \left( A+B \right)\ and\ \tan \left( A-B \right)\] to find the value of tan2A. We will be using a similar approach for tan2B also.

Complete step-by-step answer:

Now, we have been given that,

\[\begin{align}

  & \tan \left( A+B \right)=x............\left( 1 \right) \\

 & \tan \left( A-B \right)=y............\left( 2 \right) \\

\end{align}\]

Now, we have to find the values of tan2A and tan2B.

We can rewrite tan2A as,

$\tan 2A=\tan \left( \left( A+B \right)+\left( A-B \right) \right)$

Now, we will use trigonometric identity that,

$\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

So, we have,

$=\dfrac{\tan \left( A+B \right)+\tan \left( A-B \right)}{1-\tan \left( A+B \right)\tan \left( A-B \right)}$

Now, we will substitute the value from (1) and (2). So, we have,

$\tan 2A=\dfrac{x+y}{x-y}$

Now, tan2B can also be rewritten as,

$\tan 2B=\tan \left( \left( A+B \right)-\left( A-B \right) \right)$

Now, we will use trigonometric identity that,

$\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$

So, we have,

$\tan 2B=\dfrac{\tan \left( A+B \right)-\tan \left( A-B \right)}{1+\tan \left( A+B \right)\tan \left( A-B \right)}$

Now, we will substitute the value from (1) and (2). So, we have,

$\tan \left( 2B \right)=\dfrac{x-y}{1+xy}$

Note: To solve these type of questions it is important to note how we have write tan2A as $\tan \left( \left( A+B \right)+\left( A-B \right) \right)$ and used the values of \[\tan \left( A+B \right)=x\ and\ \tan \left( A-B \right)=y\] to find the values of tan2A and tan2B.