Question

The smallest positive value of $x$ (in degrees) for which $\tan (x+{100}^o)=\tan (x+{50}^o)\cdot \tan x\cdot \tan (x-{50}^o)$ is:

• A. $30$
• B. $45$
• C. $60$
• D. $55$

Answer 1

The correct option is A $30$

The relation may be written as $\frac{\tan (x+{100}^o)}{\tan (x-{50}^o)}=\tan (x+{50}^o)\tan x$

$\Rightarrow \frac{\sin (x+{100}^o)\cos (x-{50}^o)}{\sin (x-{50}^o)\cos (x+{100}^o)}=\frac{\sin (x+{50}^o)\sin x}{\cos (x+{50}^o)\cos x}$

$\Rightarrow \frac{\sin (2x+{50}^o)+\sin \left({150}^o\right)}{\sin (2x+{50}^o)-\sin \left({150}^o\right)}=\frac{\cos \left({50}^o\right)-\cos (2x+{50}^o)}{\cos \left({50}^o\right)+\cos (2x+{50}^o)}$

$\Rightarrow \frac{\sin (2x+{50}^o)}{\sin {150}^o}=\frac{-\cos {50}^o}{\cos (2x+{50}^o)}\Rightarrow \cos {50}^o+2\sin (2x+{50}^o)\cos (2x+{50}^o)=0$

$\Rightarrow \cos {50}^o+\sin (4x+{100}^o)=0\Rightarrow \cos {50}^o+\cos (4x+{10}^o)=0$

$\Rightarrow \cos (2x+{30}^o)\cos (2x-{20}^o)=0\Rightarrow x={30}^o,{55}^o$

$\Rightarrow$ The smallest value of $x={30}^o$

Hence (A) is the correct answer.