Question

If $cosx=tany,cosy=tanz$ and $cosz=tanx,$ then $sinx=2sin\theta$ where $\theta$ is (where, $x,y,z,\theta$ are acute angles)

• A. $15^o$
• B. $18^o$
• C. $22{\frac{1}{2}}^o$
• D. $75^o$

Answer 1

Answer: (B)

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