Question

The value of \[ \cot {70^ \circ } + 4\cos {70^ \circ }\] is
A. \[ \dfrac{1}{{\sqrt 3 }}\]
B. \[ \sqrt 3 \]
C. \[ 2\sqrt 3 \]
D. \[ \dfrac{1}{2}\]

Answer 1

Hint: The exact value of \[ \cot {70^ \circ }\] and \[ \cos {70^ \circ }\] in the trigonometry is not defined so we will try to convert the given function in the form where the value of the function is defined.
First by using the cofunction identity of the function convert the equation in \[ \tan {20^ \circ }\] and \[ \sin {20^ \circ }\] form and then by using sum of product trigonometric identity we will find the value of the function.

Complete step-by-step answer:
Let \[ x = \cot {70^ \circ } + 4\cos {70^ \circ }\]
Now we can also write this trigonometric equation as
 \[ x = \cot \left( {{{90}^ \circ } - {{20}^ \circ }} \right) + 4\cos \left( {{{90}^ \circ } - {{20}^ \circ }} \right)\]
Now since we know the cofunction identity of \[ \left( {\sin {{30}^ \circ } = \dfrac{1}{2}} \right)\] and \[ \cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta \] , hence we can further write the equation as
 \[
  x = \tan {20^ \circ } + 4\sin {20^ \circ } \\
   = \dfrac{{\sin {{20}^ \circ }}}{{\cos {{20}^ \circ }}} + 4\sin {20^ \circ }{\text{ }}\left( {\because \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}} \right) \\
 \]
By taking the LCM we can write
 \[
  x = \dfrac{{\sin {{20}^ \circ } + 4\sin {{20}^ \circ }\cos {{20}^ \circ }}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{\sin {{20}^ \circ } + 2\left( {2\sin {{20}^ \circ }\cos {{20}^ \circ }} \right)}}{{\cos {{20}^ \circ }}} \;
 \]
Now since we know \[ \sin 2\theta = 2\sin \theta \cos \theta \] , so we can write this equation as
 \[
  x = \dfrac{{\sin {{20}^ \circ } + 2\sin 2\left( {{{20}^ \circ }} \right)}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{\sin {{20}^ \circ } + 2\sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{\sin {{20}^ \circ } + \sin {{40}^ \circ } + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \;
 \]
Now we already know that the sum of product trigonometric identity \[ \left( {\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)} \right)\]
Hence we can further write the equation as
 \[
  x = \dfrac{{\left( {\sin {{20}^ \circ } + \sin {{40}^ \circ }} \right) + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{2\sin \dfrac{{{{60}^ \circ }}}{2}\cos \dfrac{{{{20}^ \circ }}}{2} + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{2\sin {{30}^ \circ }\cos {{10}^ \circ } + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
 \]
Now as we know the trigonometric value of \[ \left( {\sin {{30}^ \circ } = \dfrac{1}{2}} \right)\] , hence we can further write the equation as
 \[
  x = \dfrac{{2 \times \dfrac{1}{2}\cos {{10}^ \circ } + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
   \Rightarrow x = \dfrac{{\cos {{10}^ \circ } + \sin {{40}^ \circ }}}{{\cos {{20}^ \circ }}} \\
 \]
Also
  \[ x = \dfrac{{\cos {{10}^ \circ } + \cos {{50}^ \circ }}}{{\cos {{20}^ \circ }}}\left( {\because \sin \left( {{{90}^ \circ } - {{50}^ \circ }} \right) = \cos {{50}^ \circ }} \right)\]
Now we know that the sum of product trigonometric identity of \[ \left( {\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}} \right)\] , hence we can further write the equation as
 \[ x = \dfrac{{2\cos \dfrac{{{{60}^ \circ }}}{2}\cos \dfrac{{{{40}^ \circ }}}{2}}}{{\cos {{20}^ \circ }}}\]
By solving this we get
 \[ x = \dfrac{{2\cos {{30}^ \circ }\cos {{20}^ \circ }}}{{\cos {{20}^ \circ }}}\]
This is equal to
 \[ x = 2\cos {30^ \circ }\]
Now since the trigonometric values of the angle \[ \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\] , hence we can write the equation as
 \[
  x = 2 \times \dfrac{{\sqrt 3 }}{2} \\
  x = \sqrt 3 \;
 \]
Hence, the value of the given trigonometric equation \[ \cot {70^ \circ } + 4\cos {70^ \circ } = \sqrt 3 \]
Option B is correct.
So, the correct answer is “Option B”.

Note: It is worth noting down here that the sign of the trigonometric functions changes with the change in the quadratic segments like, all the trigonometric functions holds only the positive value while be in the first quadrant, only the sine and the cosec functions are positive in the second quadrant while all the other trigonometric functions are negative, only the tangent and the cot functions are positive in the third quadrant while all the other trigonometric functions are negative and only the cosine and the sec functions are positive in the fourth quadrant while all the other trigonometric functions are negative.