Question

What is the value of$4\cos {18}^o-3\sec {18}^o-2\tan {18}^o?$

• A. $0$
• B. $\frac{\sqrt{5}-1}{4}$
• C. $\frac{\sqrt{5}+1}{4}$
• D. $1$

Answer 1

Answer: (A)

$0$

Let $P={4\cos 18}^0-3{\sec 18}^0-2{\tan 18}^0$

multiply both sides with ${\cos }^2{18}^0$

then $P{\cos }^2{18}^0=4{\cos }^3{18}^0-3{\cos 18}^0-2\sin {18}^0{\cos 18}^0$

Using $\cos 3x=4{\cos }^{3}x-3\cos x$ and $\sin 2x=2\sin x\cos x$

$P{\cos }^2{18}^0=\cos \left({54}^0\right)-\sin \left({36}^0\right)=\sin \left(36\right)-\sin \left(36\right)=0$

$P{\cos }^2{18}^0=0$

Hence $P=0$