Question

How do you use the Sum and Difference Identity to find the exact value of $\tan {345^ \circ }$.

Answer 1

Hint : In order to solve this question ,split $\tan {345^ \circ }$ in to sum of angles as $\tan {345^ \circ } = \tan ({30^ \circ } + {315^ \circ })$ . Now apply the formula of sum of angles of tangent $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ considering $A$ as $30$ and $B$ as $315$ .The value of $\tan (315)$ can be find by making this a special ang as $\tan (270 + 45)$ which is equal to $- \tan 45$ as tangent is always negative in 4th quadrant .Simplifying further the formula your will get your required result.

Complete step-by-step answer :
In order the find the exact value of $\tan {345^ \circ }$ , we have to find the two angles whose either Sum or difference is ${345^ \circ }$
We only know the exact value of tangent at angles $0,{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ .
Now have to find such a combination of the above angles with ${345^ \circ }$ ,so that the sum or difference is a special angle .
If we subtract $345$ with $30$ ,we get $345 - 30 = 315$ and as we know the value of $315$ can be found by $\tan (270 + 45)$ .
$\tan (270 + 45)$ is an angle which is in the 4th quadrant .
$\tan (315) = \tan (270 + 45)$
Note that $\tan (270 + \theta ) = - \tan (\theta )$ as tangent is always negative in the 4th quadrant. So,

$$\tan (315) = \tan (270 + 45) \\\ = - \tan ({45^ \circ }) \\\ = - 1 \;$$

$\tan (315) = - 1$--------(1)
So, we can use
$\tan {345^ \circ } = \tan ({30^ \circ } + {315^ \circ })$
Using formula $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ ,
$\tan {345^ \circ } = \tan ({30^ \circ } + {315^ \circ }) \\\ = \dfrac{{\tan {{30}^ \circ } + \tan {{315}^ \circ }}}{{1 - \tan {{30}^ \circ }\tan {{315}^ \circ }}} \;$
As we know $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ and from equation (1) $\tan (315) = - 1$,our equation becomes
$= \dfrac{{\dfrac{1}{{\sqrt 3 }} + \left( { - 1} \right)}}{{1 - \left( {\dfrac{1}{{\sqrt 3 }}} \right)\left( { - 1} \right)}} \\\ = \dfrac{{\dfrac{1}{{\sqrt 3 }} - 1}}{{1 + \dfrac{1}{{\sqrt 3 }}}} \\\ = \dfrac{{\dfrac{{1 - \sqrt 3 }}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}} \\\ = \dfrac{{1 - \sqrt 3 }}{{\sqrt 3 + 1}} \;$
To remove the square term from the denominator ,multiply and divide with $\left( {\sqrt 3 - 1} \right)$
$= \dfrac{{1 - \sqrt 3 }}{{\sqrt 3 + 1}} \times \dfrac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}$
Using formula $(a - b)(a + b) = {a^2} - {b^2}$
$= \dfrac{{\sqrt 3 - 1 - 3 + \sqrt 3 }}{{3 - 1}} \\\ = \dfrac{{2\sqrt 3 - 4}}{2} \\\ = \sqrt 3 - 2 \;$
$\therefore \tan {345^ \circ } = \sqrt 3 - 2$
Therefore, the exact value of $\tan {345^ \circ }$ is equal to $\sqrt 3 - 2$
So, the correct answer is “ $\sqrt 3 - 2$ ”.

Note : 1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
3. Tangent is always positive in quadrant 1 and 3.
Formula:
$\sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \sin \left( B \right)\cos \left( A \right)$
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$(a - b)(a + b) = {a^2} - {b^2}$