Question

The smallest positive value of $x$ (in degrees) for which $\tan \left(x+100^{∘}\right)$ $=\tan \left(x+50^{∘}\right)â‹\dots \tan xâ‹\dots \tan \left(x−50^{∘}\right)$ is

• A. $25^{∘}$
• B. $82\frac{1^{∘}}{2}$
• C. $55^{∘}$
• D. $30^{∘}$

Answer 1

Answer: (D)

$\tan \left(x+100^{∘}\right)=\tan \left(x+50^{∘}\right)â‹\dots \tan xâ‹\dots \tan \left(x−50^{∘}\right)$
$⇒\frac{\tan \left(x+100^{∘}\right)}{\tan \left(x−50^{∘}\right)}=\tan \left(x+50^{∘}\right)â‹\dots \tan x$
$⇒\frac{\sin \left(x+100^{∘}\right)\cos \left(x−50^{∘}\right)}{\sin \left(x−50^{∘}\right)\cos \left(x+100^{∘}\right)}=\frac{\sin \left(x+50^{∘}\right)\sin x}{\cos \left(x+50^{∘}\right)\cos x}$
$⇒\frac{\sin \left(x+100^{∘}\right)\cos \left(x−50^{∘}\right)+\sin \left(x−50^{∘}\right)\cos \left(x+100^{∘}\right)}{\sin \left(x+100^{∘}\right)\cos \left(x−50^{∘}\right)−\sin \left(x−50^{∘}\right)\cos \left(x+100^{∘}\right)}$
$=\frac{\sin \left(x+50^{∘}\right)\sin x+\cos \left(x+50^{∘}\right)\cos x}{\sin \left(x+50^{∘}\right)\sin x−\cos \left(x+50^{∘}\right)\cos x}$
$⇒\frac{\sin \left(x+100^{∘}+x−50^{∘}\right)}{\sin \left(x+100^{∘}−x+50^{∘}\right)}=\frac{\cos \left(x+50^{∘}−x\right)}{−\cos \left(x+50^{∘}+x\right)}$
$⇒\frac{\sin \left(2x+50^{∘}\right)}{\sin \left(150^{∘}\right)}=\frac{\cos \left(50^{∘}\right)}{−\cos \left(2x+50^{∘}\right)}$
$⇒−\sin \left(2x+50^{∘}\right)\cos \left(2x+50^{∘}\right)$
$=\sin \left(90^{∘}+60^{∘}\right)\cos 50^{∘}$
$⇒−\frac{2}{2}\sin \left(2x+50^{∘}\right)\cos \left(2x+50^{∘}\right)=\cos 60^{∘}\cos 50^{∘}$
$⇒−\frac{\sin \left(4x+100^{∘}\right)}{2}=\frac{1}{2}\cos 50^{∘}$
$⇒\sin \left(4x+100^{∘}\right)=−\cos 50^{∘}$
$⇒\sin \left(4x+100^{∘}\right)=−\sin 40^{∘}$
$⇒\sin \left(4x+100^{∘}\right)=\sin \left(−40^{∘}\right)$
$=\sin \left(180^{∘}+40^{∘}\right)$
$⇒\left(4x+100^{∘}\right)=\left(180^{∘}+40^{∘}\right)$
$⇒4x+100^{∘}=220^{∘}$
$⇒4x=220^{∘}−100^{∘}=120^{∘}$
$⇒x=30^{∘}$