Question

Find the value of \[\cot {70^\circ} + 4\cos {70^\circ}\].
A. \[\sqrt 2 \]
B. \[\sqrt 3 \]
C. \[0\]
D. \[1\]

Answer 1

Hint: Simplify the given expression in terms of sine angles and cosine angles and then into cosine angles to obtain the required answer by the simple trigonometric ratio formulae and trigonometric identities. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given \[\cot {70^\circ} + 4\cos {70^\circ}\]
Using \[\tan \left( {{{90}^\circ} - x} \right) = \cot x{\text{ and }}\sin \left( {{{90}^\circ} - x} \right) = \cos x\], we have
\[
  \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \tan {20^\circ} + 4\sin {20^\circ} \\
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{\sin {{20}^\circ}}}{{\cos {{20}^\circ}}} + 4\sin {20^\circ} \\
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{\sin {{20}^\circ} + 4\sin {{20}^\circ}\cos {{20}^\circ}}}{{\cos {{20}^\circ}}} \\
\]
Using \[2\sin x\cos x = \sin 2x\], we have
\[ \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{\sin {{20}^\circ} + 2\sin {{40}^\circ}}}{{\cos {{20}^\circ}}}\]
Using \[\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)\] and \[\cos \left( { - x} \right) = \cos \left( { x} \right)\], we have
\[
\Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{2\sin \left( {\dfrac{{{{20}^\circ} + {{40}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{20}^\circ} - {{40}^\circ}}}{2}} \right) + \sin {{40}^\circ}}}{{\cos {{20}^\circ}}} \\
\Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{2\sin {{30}^\circ}\cos {{10}^\circ} + \sin {{40}^\circ}}}{{\cos {{20}^\circ}}} \\
\]
Using \[\cos \left( {{{90}^\circ} - x} \right) = \sin x{\text{ and }}\sin {30^\circ} = \dfrac{1}{2}\], we have
\[
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{2 \times \dfrac{1}{2}\cos {{10}^\circ} + \cos {{50}^\circ}}}{{\cos {{20}^\circ}}} \\
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{\cos {{10}^\circ} + \cos {{50}^\circ}}}{{\cos {{20}^\circ}}} \\
\]
Using \[\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)\], we have
\[
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{2\cos \left( {\dfrac{{{{10}^\circ} + {{50}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{10}^\circ} - {{50}^\circ}}}{2}} \right)}}{{\cos {{20}^\circ}}} \\
 \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = \dfrac{{2\cos {{30}^\circ}\cos {{20}^\circ}}}{{\cos {{20}^\circ}}} \\
   \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = 2\cos {30^\circ} \\
\]
Using \[\cos {30^\circ} = \dfrac{{\sqrt 3 }}{2}\], we have
\[ \Rightarrow \cot {70^\circ} + 4\cos {70^\circ} = 2 \times \dfrac{{\sqrt 3 }}{2} = \sqrt 3 \]
Therefore, \[\cot {70^\circ} + 4\cos {70^\circ} = \sqrt 3 \].
Thus, the correct option is B. \[\sqrt 3 \]

Note: The used trigonometric ratio formulae are \[\tan \left( {{{90}^\circ} - x} \right) = \cot x{\text{ and }}\sin \left( {{{90}^\circ} - x} \right) = \cos x\] and the trigonometric identity formulae are \[\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)\] and \[\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)\]. So, please do remember the useful formulae in solving the trigonometry problems.