Question

How do you use the angle sum or difference identity to find the exact value of \[\tan {195^ \circ }\]?

Answer 1

Hint: To solve this problem, you should know the formula for the addition and difference identity and also you must know how we can divide the $ \theta $ into the addition form or in the difference form so that we can apply the value in either one of these identities and solve this problem.

Complete step-by-step answer:
Using a different identity we are going to solve this problem. And first let us divide the $ \theta = {195^ \circ } $ into $ (180 + 15) $ and substituting the value we get,
 $ \tan ({195^ \circ }) = \tan ({180^ \circ } + {15^ \circ }) $
We know that, $ \tan ({180^ \circ } + \theta ) = \tan \theta $ and hence the above equation can be written as,
 $ \tan ({195^ \circ }) = \tan {15^ \circ } $
Now to apply the difference identity we need to divide $ \theta = {15^ \circ } $ into known $ \theta $ values, so we have to divide $ \theta = 15 $ into $ ({45^ \circ } - {30^ \circ }) $ we get,
 $ \tan {15^ \circ } = \tan ({45^ \circ } - {30^ \circ }) $
The formula for the difference identity is,
 $ \tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} $
 $ \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{\tan {{45}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{45}^ \circ }\tan {{30}^ \circ }}} $
It is mandatory to divide the value of $ \theta $ into known values and now we can able to solve this problem as we know the value of $ \tan {45^ \circ } = 1 $ and $ \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} $ . Applying the values in the formula we get,
 $
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + (1)\left( {\dfrac{1}{{\sqrt 3 }}} \right)}} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{\dfrac{{\sqrt {3}-1 }}{{\sqrt 3 }}}}{{\dfrac{{\sqrt {3}+1 }}{{\sqrt 3 }}}} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{\sqrt {3}-1 }}{{\sqrt {3}+1}} \;
  $
Taking complex conjugate we get,
 $
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{{{(\sqrt 3 - 1)}^2}}}{{{{(\sqrt 3 )}^2} - {1^2}}} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{3 - 2\sqrt 3 + 1}}{2} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{4 - 2\sqrt 3 }}{2} \\
  \tan ({45^ \circ } - {30^ \circ }) = \dfrac{{2(2 - \sqrt 3 )}}{2} \\
  \tan ({45^ \circ } - {30^ \circ }) = 2 - \sqrt 3 \\
  \tan ({45^ \circ } - {30^ \circ }) = 0.268 \;
  $
So, the correct answer is “0.268”.

Note: We can also use the addition identity to solve this problem and the formula for addition identity is, $ \dfrac{{\cot A\cot B - 1}}{{\cot A\cot B}} $ . I may give you some more hints that are $ \tan (270 - \theta ) = \cot \theta $ . With this you can solve this problem by yourself.
We can solve this problem in both methods. Before applying the formula we have to convert the value of $ \theta $ into known values, so that it is easy for us to solve this type of problem. Don’t forget to work out the alternative method.