Question

If $\cos x=\tan y,\cos y=\tan z,\cos z=\tan x$ then

• A. $\sin {18}^{\circ }$
• B. $\frac{1}{2}\sin {18}^{\circ }$
• C. $2\sin {18}^{\circ }$
• D. $3\sin {18}^{\circ }$

Answer 1

The correct option is C $2\sin {18}^{\circ }$

We have $\cos x=\tan y$

$\therefore {\cos }^{2}x={\tan }^{2}y={\sec }^{2}y-1$ ......$\left(1\right)$

But $\cos y=\tan z$

$\therefore \sec y=\cot z$

$\therefore$from $\left(1\right)$

${\cos }^{2}x={\cot }^{2}z-1$

$\therefore 1+{\cos }^{2}x={\cot }^{2}z=\frac{{\cos }^{2}z}{{\sin }^{2}z}=\frac{{\cos }^{2}z}{1-{\cos }^{2}z}$

But $\cos z=\tan x$

$\therefore 1+{\cos }^{2}x=\frac{{\tan }^{2}x}{1-{\tan }^{2}x}$

$\therefore 1+1-{\sin }^{2}x=\frac{{\sin }^{2}x{\cos }^{2}x}{1-\frac{{\sin }^{2}x}{{\cos }^{2}x}}$

$\therefore 2-{\sin }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}$

$\therefore 2-{\sin }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}$

$\therefore 2-{\sin }^{2}x=\frac{{\sin }^{2}x}{1-2{\sin }^{2}x}$

$\therefore (2-{\sin }^{2}x)(1-2{\sin }^{2}x)={\sin }^{2}x$

$\therefore 2{\sin }^{4}x-6{\sin }^{2}x+2=0$

${\sin }^{2}x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}$

But $\frac{3+\sqrt{5}}{2}>1$ whereas ${\sin }^{2}x\le 1$

${\sin }^{2}x=\frac{3-\sqrt{5}}{2}=\frac{6-2\sqrt{5}}{4}=\frac{{(\sqrt{5}-1)}^2}{4}$

$\therefore \sin x=\frac{\sqrt{5}-1}{2}=2\sin {18}^{\circ }$