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  4. If cos x=tan⁡y, cos⁡y=tan⁡z and cos z=tan⁡x, then sin x=2sin ⁡ θ where θ is (where, x,y,z,θ are acute angles)

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Q. If $cos x=tan⁡y, \, cos⁡y=tan⁡z$ and $cos z=tan⁡x,$ then $sin x=2sin ⁡ \theta $ where $\theta $ is (where, $x,y,z,\theta $ are acute angles)

NTA AbhyasNTA Abhyas 2020

Solution:

$c o s^{2}x=t a n^{2}y=s e c^{2}y-1=c o t^{2}z-1$
So, $1+c o s^{2}x=c o t^{2}z=\frac{c o s^{2} z}{s i n^{2} z}=\frac{c o s^{2} z}{1 - c o s^{2} z}$
$=\frac{t a n^{2} x}{1 - t a n^{2} x}=\frac{s i n^{2} x}{c o s^{2} x - s i n^{2} x}$
$\Rightarrow \left(2 - \left(s i n\right)^{2} x\right)=\frac{\left(s i n\right)^{2} x}{1 - 2 \left(s i n\right)^{2} x}$
$\Rightarrow \left(2 - \left(s i n\right)^{2} x\right)\left(1 - 2 \left(s i n\right)^{2} x\right)=\left(s i n\right)^{2}x$
$\Rightarrow 2s i n^{4}x-6s i n^{2}x+2=0$
$\Rightarrow s i n^{2}x=\frac{3 \pm \sqrt{9 - 4}}{2}=\frac{3 \pm \sqrt{5}}{2}$
but $ s i n^{2}x\in \left[0 , 1\right]$ so rejecting $s i n^{2}x=\frac{3 + \sqrt{5}}{2}$
we have, $\left(s i n\right)^{2}x=\frac{3 - \sqrt{5}}{2}=\left(\frac{\sqrt{5} - 1}{2}\right)^{2 }$
$\Rightarrow sin x=\frac{\sqrt{5} - 1}{2}\Rightarrow 2sin ⁡ \theta =\frac{\sqrt{5} - 1}{2}$
$\Rightarrow sin \theta =\frac{\sqrt{5} - 1}{4}\Rightarrow \theta =18^{o}$