Q65 / 160 · AP EAMCET 2008 · Mathematics
If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$, then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to
- A$0$
- Correct answerB$1$
- C$2$
- D$3$
Solution
Given that, $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$
$$
\begin{aligned}
& \therefore \tan (A+B+C) \\
& =\frac{\left[\begin{array}{c}
\tan A+\tan B+\tan C \\
-\tan A \tan B \tan C
\end{array}\right]}{\left[\begin{array}{c}
1-\tan A \tan B-\tan B \tan C \\
-\tan C \tan A
\end{array}\right]} \\
& \Rightarrow \tan \left(90^{\circ}\right) \\
& =\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A} \\
& \Rightarrow \quad \tan A \tan B+\tan B \tan C \\
& +\tan C \tan A=1 \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'B'.