Question

Solve: $4\cos {18}^o-3\sec {18}^o-2\tan {18}^o$

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

Answer 1

Answer: (A)

$0$

$LetA=4\cos {18}^o-3\sec {18}^0-2\tan {18}^o\Rightarrow A\left({\cos }^2{18}^o\right)=4{\cos }^3{18}^o-3\cos {18}^o-2\sin {18}^o.\cos {18}^o\because \cos 3x=4{\cos }^{3}x-3\cos xand\sin 2x=2\sin x.\cos x\therefore A\left({\cos }^2{18}^o\right)=\cos {54}^o-\sin {36}^o\Rightarrow A\left({\cos }^2{18}^o\right)=\cos ({90}^o-{36}^o)-\sin {36}^o\Rightarrow A\left({\cos }^2{18}^o\right)=\sin {36}^o-\sin {36}^o\Rightarrow A\left({\cos }^2{18}^o\right)=0\Rightarrow A=0$