Question

The value of $\tan 27\tan 31+\tan 32\tan 31+\tan 31\tan 27$ is

• A. 0
• B. 1
• C. 2
• D. cannot be determined

Answer 1

The correct option is B 1

From given, we have,

$27+32+31=90$

$S=\tan A\tan B+\tan B\tan C+\tan C\tan A$.......(1)

$\Rightarrow S=\tan A\tan B+\tan C(\tan A+\tan B)$

where $A+B+C=90$, and hence, $C=90-(A+B)$

$\tan C=\tan (90-A-B)=\cot (A+B)=\frac{1}{\tan (A+B)}$

$\tan C=\frac{1-\tan A\tan B}{\tan A+\tan B}$

$\tan C(\tan A+\tan B)=1-\tan A\tan B$....(2)

from (1) and (2)

$S=\tan A\tan B+\tan C(\tan A+\tan B)$

$=\tan A\tan B+(1-\tan A\tan B)$

$=1$