If $ A=B=60^{\circ} $, verify that$ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} $

Given:

\( A=B=60^{\circ} \)

To do:

We have to verify that \( \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \).

Solution:  

We know that,

$\tan 60^{\circ}=\sqrt3$

Let us consider LHS,

$\tan (A-B)=\tan (60^{\circ}-60^{\circ})$

$=\tan 0^{\circ}$      

$=0$      (Since $\tan 0^{\circ}=0$)

Let us consider RHS,

$\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{\tan 60^{\circ}-\tan 60^{\circ}}{1+\tan 60^{\circ} \tan 60^{\circ}}$

$=\frac{\sqrt{3} -\sqrt{3}}{1+\left(\sqrt{3}\right)\left(\sqrt{3}\right)}$

$=0$

LHS $=$ RHS

Hence proved.   

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Updated on: 10-Oct-2022

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