Question

If $\frac{\tan {26}^o+\tan {19}^o}{1-\tan {26}^o\tan {19}^o}=\cos {60}^o$, what is the value of $x$?

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. $\sqrt{3}$

Answer 1

The correct option is C $2$

$\frac{\tan {26}^{\circ }+\tan {19}^{\circ }}{1-\tan {26}^{\circ }\tan {19}^{\circ }}=x\cos {60}^{\circ }$

$\Rightarrow \tan ({26}^{\circ }+{19}^{\circ })=x\times \frac{1}{2}$ $[\because \tan A+B=\frac{\tan A+\tan B}{1-\tan A\tan B}]$

$\Rightarrow \tan {45}^{\circ }=\frac{x}{2}$

$\Rightarrow 1=\frac{x}{2}$

$\Rightarrow x=2$

Hence, the answer is $2.$