Question

If x, y and z are acute angles and $\cos x=\tan y,\cos y=\tan z$ and $\cos z=\tan x$, then $1234\sin x+1342\sin y+1423\sin z=k\sin {18}^o$ where $k$ is equal to

Answer 1

Given

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

$\cos y=\tan z\cdots \left(2\right)$

$\cos z=\tan x\cdots \left(3\right)$

Let $x=y=z$

$⟹\cos x=\tan x$

$⟹\cos x=\frac{\sin x}{\cos x}$

$⟹\sin x={\cos }^{2}x$

$⟹\sin x=1-{\sin }^{2}x$

$⟹{\sin }^{2}x+\sin x-1=0$

$⟹\sin x=\frac{-1\pm \sqrt{1+4}}{2}$

$⟹\sin x=\frac{-1\pm \sqrt{5}}{2}$

$⟹\sin x=2\sin {18}^{\circ }$

$⟹1234\sin x+1342\sin x+1423\sin x$

$⟹1234\left(2\sin 18\right)+1342\left(2\sin 18\right)+1423\left(2\sin 18\right)$

$⟹7998\sin 18$

$⟹k=7998$