Question

$\tan {100}^{\circ }+\tan {125}^{\circ }+\tan {100}^{\circ }\tan {125}^{\circ }$ is equal to

• A. $0$
• B. $\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

The correct option is D $1$

we know that $\tan (\alpha +\beta )=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$

Putting $\alpha ={100}^o\beta ={125}^o$

$\tan ({100}^o+{125}^o)=\tan {225}^o=\frac{\tan {100}^o+\tan {125}^o}{1-\tan {100}^o\times \tan {125}^o}\Rightarrow 1=\frac{\tan {100}^o+\tan {125}^o}{1-\tan {100}^o\times \tan {125}^o}$

$\Rightarrow 1-\tan {100}^o\times \tan {125}^o=\tan {100}^o+\tan {125}^o$

$\Rightarrow \tan {100}^o\times \tan {125}^o+\tan {100}^o+\tan {125}^o=1$

Therefore, Answer is $1$