Question

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

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

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

Note – In such types of questions first multiply the equation by ${\cos ^2}{18^0}$ 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 $\cos \theta = \sin \left( {90 - \theta } \right)$ and again simplify then we will get the required answer.