How do you find the exact functional value $\tan 345$ using the cosine sum or difference identity?

Hint: Apply the sum and difference Trigonometric identities to find the exact functional value.
Here write $\tan 345$ into two parts mean $\tan \left( 145+180 \right)$
Use the formula for $\sin \left( a-b \right)=\sin a.\cos b-\sin b.\cos a$
For solving the problem and $\cos \left( a-b \right)=\cos a.\cos b+\sin a.\sin b$

Complete step by step solution:As per given problem trigonometric function is $\tan 345$ Now, we have to split $'345'$ into two parts in the form of addition by adding any two numbers whose addition should be $345{}^\circ $
$\therefore \tan 345{}^\circ =\tan \left( 165{}^\circ +180{}^\circ \right)$
$=\tan \left( 165{}^\circ \right)$
Neglecting $\tan \left( 180{}^\circ \right)$ because the solution of $\tan \left( 180{}^\circ \right)=\dfrac{\sin \left( 180{}^\circ \right)}{\cos \left( 180{}^\circ \right)}$
It comes from the trigonometric identity,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }=0$
Value of $\sin \left( 180{}^\circ \right)$ is equal to $0$ and value of $\cos \left( 180{}^\circ \right)$ is $-1$
$\tan \left( 180{}^\circ \right)=\dfrac{0}{-1}$
$\tan \left( 180{}^\circ \right)=0$
Therefore neglect $\tan \left( 180{}^\circ \right)$ and solve $\tan \left( 165{}^\circ \right)$
$\tan \left( 165{}^\circ \right)=-15\tan \text{degree}=-\tan 15{}^\circ $
Here again apply the formula of $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Therefore,
$\tan 345{}^\circ =-\tan 15{}^\circ =\dfrac{-\sin 15{}^\circ }{\cos 15{}^\circ }...(ii)$
From the above ${{1}^{st}}$ equation, firstly solve the numerator,
$\therefore \sin \left( 15{}^\circ \right)$
Split $15{}^\circ $ into two parts whose solution will be $15{}^\circ $ after subtracting.
$\therefore \sin \left( 15{}^\circ \right)=\sin \left( 45{}^\circ -30{}^\circ \right)$
This $\sin \left( 45{}^\circ -30{}^\circ \right)$ is applicable to the trigonometric identity which is.
$\sin \left( a-b \right)=\sin a.\cos b-\sin b.\cos a$
Here, $a=45{}^\circ $
$b=30{}^\circ $
Apply trigonometric identity.
$\sin \left( 15{}^\circ \right)=\left( \sin 45{}^\circ -30{}^\circ \right)=\sin 45.\cos 30-\sin 30.\cos 45$
$=\left( \dfrac{\sqrt{2}}{2} \right).\left( \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{2} \right).\left( \dfrac{\sqrt{2}}{2} \right)$
This above value is written from a trigonometric function table.
Simplify the above equation.
$=\left( \dfrac{\sqrt{6}}{4} \right)-\left( \dfrac{\sqrt{2}}{4} \right)$
$\therefore \sin \left( 15{}^\circ \right)=\dfrac{\sqrt{6}-\sqrt{2}}{4}$
Solve denominator of ${{1}^{st}}$ equation, which is $\cos \left( 15{}^\circ \right)$ which can also be written as,
$\cos \left( 45{}^\circ -30{}^\circ \right)$
Here, apply $\cos \left( a-b \right)=\cos a.\cos b+\sin a.\sin b$ trigonometric identity.
$\cos \left( 15{}^\circ \right)=\cos \left( 45{}^\circ -30{}^\circ \right)=\cos 45{}^\circ .\cos 30{}^\circ +\sin 45{}^\circ -\sin 30{}^\circ $
$=\left( \dfrac{\sqrt{2}}{2} \right).\left( \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{\sqrt{2}}{2} \right).\left( \dfrac{1}{2} \right)$
Multiplying and adding above value.
$=\left( \dfrac{\sqrt{6}}{4} \right)+\left( \dfrac{\sqrt{2}}{4} \right)$
$\cos \left( 15{}^\circ \right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}$
Denominators can be written any $'1'$ time as the denominator is common.
From equation ${{1}^{st}}$
$\tan 345{}^\circ =-\tan 15{}^\circ =\dfrac{-\sin 15{}^\circ }{\cos 15{}^\circ }$
$=\dfrac{\sqrt{6}-\sqrt{2}}{4}$
$=\dfrac{\sqrt{6}+\sqrt{2}}{4}$
$=\dfrac{\sqrt{6}-\sqrt{2}}{4}\times \dfrac{4}{\sqrt{6}+\sqrt{2}}$
Here, denominator comes in multiplication with numerator and as per the rule denominator will be reciprocal and $'4'$ in division and $'4'$ in multiplication will get canceled.
Therefore,
$\tan 345{}^\circ =\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Additional Information:
The main sum and difference trigonometric identities are following.
(1) $\cos \left( a-b \right)=\cos a.\cos b+\sin a.\sin b$
(2) $\cos \left( a+b \right)=\cos a.\cos b-\sin a.\sin b$
(3) $\sin \left( a+b \right)=\sin a.\cos b+\sin b.\cos a$
(4) $\sin a\left( a+b \right)=\sin a.\cos b+\sin b.\cos a$
(5) $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}$
(6) $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a.\tan b}$
Here is one example which is applicable to sum and difference trigonometric identities.
Here is one example which is applicable to sum and difference trigonometric identities.
(1) find $\sin 2a$
$=\sin \left( a+a \right)$ split into two parts, Apply the sum or difference trigonometric identity.
i.e. $\sin \left( a+b \right)=\sin a.\cos b+\sin b.\cos a$
$\because \sin 2a=\sin \left( a+a \right)$
$=\sin a.\cos a+\sin a.\cos a$
$\sin 2a=2.\left( \sin a.\cos a \right)$

Note:
Always remember while applying sum and difference identities in the function split the value of degree into two points for saving sometimes.
Apply the trigonometric identities such as $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\sin \theta =\dfrac{1}{\cos \theta }$
Where applicable,
Remember the value of trigonometric function angle such as $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$
$\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$, etc.