Question

$\tan {20}^o+\tan {25}^o+\tan {20}^o\cdot \tan {25}^o=$_____

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

Answer 1

Answer: (B)

$1$

$\tan ({20}^{\circ }+{25}^{\circ })=\tan {45}^{\circ }$

$⟹\frac{\tan {20}^{\circ }+\tan {25}^{\circ }}{1-\tan {20}^{\circ }\tan {25}^{\circ }}=1$

$⟹\tan {20}^{\circ }+\tan {25}^{\circ }=1-\tan {20}^{\circ }\tan {25}^{\circ }$

$⟹\tan {20}^{\circ }+\tan {25}^{\circ }+\tan {20}^{\circ }\tan {25}^{\circ }=1$