Question

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

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
• C. Assertion is correct but Reason is incorrect.
• D. Assertion is incorrect but Reason is correct.

Answer 1

The correct option is D Assertion is incorrect but Reason is correct.

In any triangle ABC, $A+B+C=\pi$
$\Rightarrow A=\pi -(B+C)$
$\tan A=-\tan (B+C)$
$\Rightarrow \tan A=-\frac{\tan B+\tan C}{1-\tan B\tan C}$
$\Rightarrow \tan A=\frac{\tan B+\tan C}{\tan B\tan C-1}$ ...(1)
$\Rightarrow$ statement-2 is correct.
Since B and C are in first quadrant
$\tan B,\tan C>0$
But ,since A is obtuse i.e. A is in second quadrant
$\Rightarrow \tan A<0$
Now, by (1), $\frac{\tan B+\tan C}{\tan B\tan C-1}<0$
Since B and C are in first quadrant
$\tan B,\tan C>0$, so denominator should be less than 0
So $\tan B\tan C<1$and thus statement-1 is false.