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How do you use the sum and difference identities to find the exact value of \[\tan ({105^0})\]?

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Last updated date: 20th Jun 2024
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Answer
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Hint: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions. We know the sum formula for tangent that is \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\]. We can split 105 as a sum of 60 and 45.

Complete step-by-step solution:
Given.
\[\tan ({105^0})\]
Since we can express 105 as the sum of 60and 45. That is \[{105^0} = {45^0} + {60^0}\].
Then,
\[\Rightarrow \tan ({105^0}) = \tan \left( {{{45}^0} + {{60}^0}} \right)\]
\[ \Rightarrow \tan \left( {{{45}^0} + {{60}^0}} \right)\]
Applying the sum formula of tangent. That is \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\], where \[A = {45^0}\]and \[B = {60^0}\].
\[\Rightarrow \dfrac{{\tan {{45}^0} + \tan {{60}^0}}}{{1 - \tan {{45}^0}.\tan {{60}^0}}}\]
We know \[\tan {45^0} = 1\] and \[\tan {65^0} = \sqrt 3 \]. Substituting we have,
\[
   \Rightarrow \dfrac{{1 + \sqrt 3 }}{{1 - \left( {1.\sqrt 3 } \right)}} \\
   \Rightarrow \dfrac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }} \\
 \]
We can stop here. If we want we can rationalize the obtained solution.
\[ \Rightarrow \dfrac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }} \times \dfrac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}\]
We have the identities \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and \[{a^2} - {b^2} = (a + b)(a - b)\], using these we have
\[
   \Rightarrow \dfrac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{{{1^2} - {{\left( {\sqrt 3 } \right)}^2}}} \\
   \Rightarrow \dfrac{{{1^2} + {{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3 }}{{1 - 3}} \\
   \Rightarrow \dfrac{{1 + 3 + 2\sqrt 3 }}{{ - 2}} \\
   \Rightarrow \dfrac{{4 + 2\sqrt 3 }}{{ - 2}} \\
 \]
Taking 2 common and simplifying we have
\[
   \Rightarrow \dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{ - 2}} \\
   \Rightarrow - \left( {2 + \sqrt 3 } \right) \\
 \]
Thus we have \[\tan ({105^0}) = - \left( {2 + \sqrt 3 } \right)\] .

Note: Remember the formula sine and cosine addition and subtract formula well. In case id sine if we have cosine we will use the formula \[\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)\]. Similarly we have \[\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b)\] and \[\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)\]. Depending on the angle we use the required formulas. Also remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.