Question

Prove that: $\dfrac{\tan \left( A+B \right)}{\cot \left( A-B \right)}=\dfrac{{{\tan }^{2}}A-{{\tan }^{2}}B}{1-{{\tan }^{2}}A{{\tan }^{2}}B}$.

Answer 1

Hint: We will be using the concept of trigonometric ratio to solve the problem. We will first take LHS of the equation and apply the trigonometric identity that \[\cot \theta =\dfrac{1}{\tan \theta }\], then we will apply the trigonometric identity that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, then we will use algebraic identity like $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to further simplify the problem.

Complete step-by-step answer:

Now, we have in LHS as,

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

We know that \[\cot \theta =\dfrac{1}{\tan \theta }\]. Therefore, using this we have,

\[\begin{align}

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

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

\end{align}\]

Now, we know the trigonometric identities that

$\begin{align}

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

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

\end{align}$

Now, we will substitute these values in (1). So, we have,

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

Now, we will solve the numerator and denominator respectively. So, we have,

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

Now, we will apply the trigonometric identity that,

$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$

So, we have now that,

$=\dfrac{{{\tan }^{2}}A-{{\tan }^{2}}B}{1-{{\tan }^{2}}A{{\tan }^{2}}B}$

Since, LHS = RHS. Hence, we have prove that,

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

Note: To solve these types of questions it is important to note that we have first converted the whole LHS in terms of $\tan \theta $, then we expand tan (A + B) and tan (A – B) to get the same expression as that of the right hand side.