Q. If $A = 35{^\circ},\ B = 15{^\circ}$ and $C = 40{^\circ}$, then $\tan{A.\tan B} + \tan{B.\tan{C + \tan C}}.\tan A$ is equal to
A) 0
B) 1
C) 2
D) 3
Explanation
(b)
$\tan{(A + B + C) = \frac{\lbrack\tan{A + \tan{B + \tan{C - \tan{A\tan{B\tan C}}}}}\rbrack}{\lbrack 1 - \tan{A\tan{B - \tan{B\tan{C - \tan{C\tan A}}}}}\rbrack}}$
$\Rightarrow \ \ \tan{(90{^\circ})}$
$= \frac{\tan{A + \tan{B + \tan{C - \tan{A\tan{B\tan C}}}}}}{1 - \tan{A\tan{B - \tan{B\tan{C - \tan{C\tan A}}}}}}$
$\Rightarrow \ \tan{A\tan{B + \tan{B\tan{C + \tan{C\tan{A = 1}}}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.