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} $