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# How do you prove $\tan \left( 180{}^\circ +a \right)=\tan \left( a \right)$

Last updated date: 25th Jun 2024
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Hint: Here in the given question we have to prove the given following for that we must use here the fundamental trigonometric identities for proving the given term. As the given turn is in $\tan \left( a+b \right)$ form. We have the identity of $\tan \left( a+b \right)=\tan a+\tan b$
$=1-\tan a.\tan b$
To prove the given term. There are so many trigonometric identities which are used in various problems. We have to remember all their identities.

Complete step by step solution:
Here we have,
$\tan \left( 180{}^\circ +a \right)$
But we have to prove here $\tan \left( 180{}^\circ +a \right)=\tan \left( a \right)$
So, as we seen the problem it shows that it is a same as the $\tan \left( a+b \right)$
This is the trigonometric identity. There are also various identities.
The trigonometric Identity for
$\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a-\tan b}$
So, Here $b=180{}^\circ$ and $\tan 180{}^\circ =0$
You have to remembers one point that the period of $\sin \left( t \right)$ and $\cos \left( t \right)$ is $2\pi =360{}^\circ$ and
In case of $\tan \left( t \right)$ It is $\pi =180{}^\circ$
Hence,
It is proved that $\tan \left( 180{}^\circ +a \right)=\tan \left( a \right)$