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  4. The smallest positive value of x (in degrees) for which tan (x+100°) = tan (x+50°) ⋅ tan x ⋅ tan (x-50°) is

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Q. The smallest positive value of $x$ (in degrees) for which $\tan \left(x+100^{\circ}\right)$ $=\tan \left(x+50^{\circ}\right) \cdot \tan x \cdot \tan \left(x-50^{\circ}\right)$ is

AP EAMCETAP EAMCET 2019

Solution:

$\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \cdot \tan x \cdot \tan \left(x-50^{\circ}\right)$
$\Rightarrow \frac{\tan \left(x+100^{\circ}\right)}{\tan \left(x-50^{\circ}\right)}=\tan \left(x+50^{\circ}\right) \cdot \tan x$
$\Rightarrow \frac{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)}{\sin \left(x-50^{\circ}\right) \cos \left(x+100^{\circ}\right)}=\frac{\sin \left(x+50^{\circ}\right) \sin x}{\cos \left(x+50^{\circ}\right) \cos x}$
$\Rightarrow \frac{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)+\sin \left(x-50^{\circ}\right) \cos \left(x+100^{\circ}\right)}{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)-\sin \left(x-50^{\circ}\right) \cos \left(x+100^{\circ}\right)}$
$=\frac{\sin \left(x+50^{\circ}\right) \sin x+\cos \left(x+50^{\circ}\right) \cos x}{\sin \left(x+50^{\circ}\right) \sin x-\cos \left(x+50^{\circ}\right) \cos x}$
$\Rightarrow \frac{\sin \left(x+100^{\circ}+x-50^{\circ}\right)}{\sin \left(x+100^{\circ}-x+50^{\circ}\right)}=\frac{\cos \left(x+50^{\circ}-x\right)}{-\cos \left(x+50^{\circ}+x\right)}$
$ \Rightarrow \frac{\sin \left(2 x+50^{\circ}\right)}{\sin \left(150^{\circ}\right)}=\frac{\cos \left(50^{\circ}\right)}{-\cos \left(2 x+50^{\circ}\right)}$
$\Rightarrow -\sin \left(2 x+50^{\circ}\right) \cos \left(2 x+50^{\circ}\right)$
$=\sin \left(90^{\circ}+60^{\circ}\right) \cos 50^{\circ}$
$\Rightarrow -\frac{2}{2} \sin \left(2 x+50^{\circ}\right) \cos \left(2 x+50^{\circ}\right)=\cos 60^{\circ} \cos 50^{\circ}$
$\Rightarrow -\frac{\sin \left(4 x+100^{\circ}\right)}{2}=\frac{1}{2} \cos 50^{\circ}$
$\Rightarrow \sin \left(4 x+100^{\circ}\right)=-\cos 50^{\circ}$
$\Rightarrow \sin \left(4 x+100^{\circ}\right)=-\sin 40^{\circ}$
$\Rightarrow \sin \left(4 x+100^{\circ}\right)=\sin \left(-40^{\circ}\right)$
$=\sin \left(180^{\circ}+40^{\circ}\right)$
$\Rightarrow \left(4 x+100^{\circ}\right)=\left(180^{\circ}+40^{\circ}\right)$
$\Rightarrow 4 x+100^{\circ}=220^{\circ}$
$\Rightarrow 4 x=220^{\circ}-100^{\circ}=120^{\circ}$
$\Rightarrow x=30^{\circ}$