Question

If $cosx=tany,coty=tanz,cotz=tanx$ then $sinx=$

• A. $\sin x=\sin y=\sin z=\sin {24}^{\circ }$
• B. $\sin x=\sin y=\sin z=\sin {18}^{\circ }$
• C. $\sin x=\sin y=\sin z=\sin {36}^{\circ }$
• D. $\sin x=\sin y=\sin z=\sin {44}^{\circ }$

Answer 1

The correct option is B $\sin x=\sin y=\sin z=\sin {18}^{\circ }$

$\cos x=\tan y$ .....$\left(1\right)$

$\cot y=\tan z$ ......$\left(2\right)$

$\cot z=\tan x$ .......$\left(3\right)$

$\Rightarrow \cos x\tan z\cot z=\tan y\cot y\tan x$ from $\left(1\right),\left(2\right)$ and $\left(3\right)$

$\Rightarrow \cos x\times 1=1\times \tan x$

$\Rightarrow \cos x=\frac{\sin x}{\cos x}$

$\Rightarrow {\cos }^{2}x=\sin x$

$\Rightarrow 1-{\sin }^{2}x-\sin x=0$

$\Rightarrow -{\sin }^{2}x-\sin x+1=0$

$\Rightarrow {\sin }^{2}x+\sin x-1=0$

$\Rightarrow \sin x=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$

Now $\sin {18}^{\circ }$ is positive, as ${18}^{\circ }$ lies in first quadrant.

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

Hence $\sin x=\sin y=\sin z=\sin {18}^{\circ }$