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# The smallest positive value of $x$ (in degrees) for which $\tan ( x+100 ^{\circ}$$)=\tan (x+50^{\circ}) \times \tan x \times \tan (x-50^{\circ}) (a)\;60^{\large\circ}\qquad(b)\;90^{\large\circ}\qquad(c)\;45^{\large\circ}\qquad(d)\;30^{\large\circ} Can you answer this question? ## 1 Answer 0 votes We have \tan (x+100^{\large\circ})=\tan (x+50^{\large\circ})\times \tan x\times \tan (x-50^{\large\circ}) \Rightarrow \large\frac{\tan (x+100^{\large\circ})}{\tan (x-50^{\large\circ})}$$=\tan(x+50^{\large\circ})\tan x$
$\Rightarrow \large\frac{\sin(x+100^{\large\circ})\cos(x-50^{\large\circ})}{\cos(x+100^{\large\circ})\sin(x-50^{\large\circ})}=\frac{\sin(x+50^{\large\circ})\sin x}{\cos(x+50^{\large\circ})\cos x}$
$\Rightarrow \large\frac{\sin(2x+50^{\large\circ})}{\sin 150^{\large\circ}}=\frac{\cos 50^{\large\circ}}{-\cos(2x+50^{\large\circ})}$
[By applying componendo & dividendo]
$\Rightarrow \sin(2x+50^{\large\circ})\cos(2x+50^{\large\circ})=-\sin 150^{\large\circ}\cos 50^{\large\circ}$
$\Rightarrow 2\sin(2x+50^{\large\circ})\cos(2x+50^{\large\circ})=-\cos 50^{\large\circ}$
$\Rightarrow \sin(4x+100^{\large\circ})=\sin(270^{\large\circ}-50^{\large\circ})$
$\Rightarrow \sin(4x+100^{\large\circ})=\sin 220^{\large\circ}$
$\Rightarrow 4x+100^{\large\circ}=220^{\large\circ}$
$\Rightarrow 4x=220^{\large\circ}-100^{\large\circ}$
$\Rightarrow 4x=120^{\large\circ}$
$\Rightarrow x=\large\frac{120^{\large\circ}}{4}$
$x=30^{\large\circ}$
Hence (d) is the correct option.
edited Mar 22, 2014