Question

Determine the smallest positive value of x (in degrees) for which $\tan (x+{100}^o)=\tan (x+{50}^o)\left(\tan x\right)\tan (x-{50}^o)$

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

The correct option is A ${30}^o$

Given $\tan (x+{100}^o)=\tan (x+{50}^o)\tan x\tan (x-{50}^o)$
$\frac{\tan (x+{100}^o)}{\tan (x-{50}^o)}=\tan (x+{50}^o)\tan x$
$\Rightarrow \frac{\sin (x+{100}^o)\cos (x-{50}^0)}{\cos (x+{100}^o)\sin (x-{50}^0)}=\frac{\sin (x+{50}^o)\sin x}{\cos (x+{50}^o)\cos x}$
Applying componendo and dividendo , we get
$\frac{\sin (2x+{50}^o)}{\sin {150}^0}=\frac{-\cos {50}^o}{\cos (2x+{50}^o)}$
$\sin (2x+{50}^o)\cos (2x+{50}^o)=-\sin {150}^0\cos {50}^o$
$\Rightarrow \sin (4x+{100}^o)=-\sin {40}^0$
$\sin (4x+{100}^o)=\sin (-{40}^0)$
$4x+{100}^o=n\pi -{(-1)}^n{40}^0$
Substitute $n=1$, smallest positive value of x is given by
$4x={120}^0$
$\Rightarrow x={30}^0$