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Question: What is the value of \(4\cos {18^0} - 3\sec {18^0} - 2\tan {18^0}\). \( {\text{a}}{\text{. 0}}...

What is the value of 4cos⁡180−3sec⁡180−2tan⁡1804\cos {18^0} - 3\sec {18^0} - 2\tan {18^0}.
a. 0 b. 5−14 c. 5+14 d. 1  {\text{a}}{\text{. 0}} \\\ {\text{b}}{\text{. }}\dfrac{{\sqrt 5 - 1}}{4} \\\ {\text{c}}{\text{. }}\dfrac{{\sqrt 5 + 1}}{4} \\\ {\text{d}}{\text{. 1}} \\\

Explanation

Solution

Hint – In this question apply some basic properties of trigonometric identities such as cos⁡3θ=(4cos⁡3θ−3cos⁡θ), sin⁡2θ=2sin⁡θcos⁡θ\cos 3\theta = \left( {4{{\cos }^3}\theta - 3\cos \theta } \right),{\text{ }}\sin 2\theta = 2\sin \theta \cos \theta , to reach the solution of the problem.

Let,
x=4cos⁡180−3sec⁡180−2tan⁡180x = 4\cos {18^0} - 3\sec {18^0} - 2\tan {18^0}
In above equation multiply both sides by cos⁡2180{\cos ^2}{18^0} , we have
x.cos⁡2180=(4cos⁡180−3sec⁡180−2tan⁡180)cos⁡2180x.{\cos ^2}{18^0} = \left( {4\cos {{18}^0} - 3\sec {{18}^0} - 2\tan {{18}^0}} \right){\cos ^2}{18^0}
Now as we know that sec⁡θ.cos⁡θ=1, tan⁡θ=sin⁡θcos⁡θ\sec \theta .\cos \theta = 1,{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}, so use this property and simplify the above equation we have,
x.cos⁡2180=(4cos⁡3180−3sec⁡180cos⁡180cos⁡180−2sin⁡180cos⁡180cos⁡2180) x.cos⁡2180=4cos⁡3180−3cos⁡180−2sin⁡180cos⁡180 \begin{gathered} x.{\cos ^2}{18^0} = \left( {4{{\cos }^3}{{18}^0} - 3\sec {{18}^0}\cos {{18}^0}\cos {{18}^0} - 2\dfrac{{\sin {{18}^0}}}{{\cos {{18}^0}}}{{\cos }^2}{{18}^0}} \right) \\\ x.{\cos ^2}{18^0} = 4{\cos ^3}{18^0} - 3\cos {18^0} - 2\sin {18^0}\cos {18^0} \\\ \end{gathered}
Now as we all know cos⁡3θ=(4cos⁡3θ−3cos⁡θ), sin⁡2θ=2sin⁡θcos⁡θ\cos 3\theta = \left( {4{{\cos }^3}\theta - 3\cos \theta } \right),{\text{ }}\sin 2\theta = 2\sin \theta \cos \theta , so use this property in above equation we have,
x.cos⁡2180=cos⁡(3×18)0−sin⁡(2×180) x.cos⁡2180=cos⁡(54)0−sin⁡(360) \begin{gathered} x.{\cos ^2}{18^0} = \cos {\left( {3 \times 18} \right)^0} - \sin \left( {2 \times {{18}^0}} \right) \\\ x.{\cos ^2}{18^0} = \cos {\left( {54} \right)^0} - \sin \left( {{{36}^0}} \right) \\\ \end{gathered}
Now we know that cos⁡θ=sin⁡(90−θ)\cos \theta = \sin \left( {90 - \theta } \right), so use this property in above equation we have,
⇒x.cos⁡2180=sin⁡(90−54)0−sin⁡(360) ⇒x.cos⁡2180=sin⁡(36)0−sin⁡(360)=0 ⇒x=0cos⁡2180=0  \Rightarrow x.{\cos ^2}{18^0} = \sin {\left( {90 - 54} \right)^0} - \sin \left( {{{36}^0}} \right) \\\ \Rightarrow x.{\cos ^2}{18^0} = \sin {\left( {36} \right)^0} - \sin \left( {{{36}^0}} \right) = 0 \\\ \Rightarrow x = \dfrac{0}{{{{\cos }^2}{{18}^0}}} = 0 \\\
So this is the required answer.
Hence, option (a) is correct.

Note – In such types of questions first multiply the equation by cos⁡2180{\cos ^2}{18^0} in both sides of the equation, then convert R.H.S part of the question into standard formulas of trigonometric identities which is stated above and simplify then use the property that cos⁡θ=sin⁡(90−θ)\cos \theta = \sin \left( {90 - \theta } \right) and again simplify then we will get the required answer.