Question

Assertion :Statement 1:If $x+y+z=xyz,$ then at most one of the numbers can be negative. Reason: Statement 2: In a triangle ABC, $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ ,there can be at most one obtuse angle in a triangle.

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1.
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE.

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.

We know from conditional identities, in trigonometry that

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

Assuming

$x=\tan A,y=\tan B,z=\tan C$

we get

$x+y+z=xyz$

Since

$A,B,C$ are angles of a triangle, and sum of all the interior angles of the is ${180}^0$ therefore if one of the angle is obtuse $(\tan \theta <0)$ then the sum of other must be acute, thus making those remaining two angles acute.

Hence atmost one angle can be obtuse.

Hence atmost one $\tan \theta$ can be negative.

Hence out of $x,y,z$ atmost one can be negative.