Question

If $\cos x=\tan y$; $\cos y=\tan z$; $\cos z=\tan x$ then the value of $\sin x$ is

• A. $2\cos {18}^0$
• B. $\cos {18}^0$
• C. $\sin {18}^0$
• D. $2\sin {18}^0$

Answer 1

The correct option is D $2\sin {18}^0$

we have,

$\cos x=\tan y$

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

$\cos y=\tan z\Rightarrow \sec y=\cot z$

from (1 )$

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

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

But, $\cos z=\tan x$

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

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

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

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

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

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

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