Question

Assertion :Statement 1: If $A,B,C$ are the angles of a triangle such that angle $A$ is obtuse,then $\tan B\tan C>1$ Reason: Statement 2: In any triangle, $\tan A=\frac{\tan B+\tan C}{\tan B\tan C-1}$

• 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 D Statement 1 is false and statement 2 is true

Let $A,B,C$ be the angles of triangle

$\therefore A+B+C=\pi$

$\tan (B+C)=\tan (\pi -A)$

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

$\Rightarrow \tan A=\frac{\tan B+\tan C}{\tan B\tan C-1}$
Since A is an obtuse angle.

$\therefore A>\frac{\pi }{2}\Rightarrow \frac{\tan B+\tan C}{\tan B\tan C-1}>\tan \frac{\pi }{2}$

$\Rightarrow \frac{\tan B+\tan C}{\tan B\tan C-1}>\frac{1}{0}$

$\therefore \tan B\tan C<1$
Ans: D