In a triangle ABC, if angle C is obtuse and angles A and B are given by roots of the equation $t a n^{2} x + p t a n x + q = 0$ , then the value of q is

greater than 1

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less than 1

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equal to 1

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Solution

The correct option is B
less than 1

We have $A + B = π - C$

$= t a n (A + B) = - t a n C$

$= \frac{t a n A + t a n B}{1 - t a n A . t a n B} > 0$ $[∵ t a n A > 0, t a n B > 0, t a n C < 0]$

$= t a n A . t a n B < 1 \Rightarrow q < 1$

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In a triangle ABC, if angle C is obtuse and angles A and B are given by roots of the equation tan2x+p tanx+q=0, then the value of q is

The correct option is B less than 1 We have A+B=π−C =tan(A+B)=−tanC =tan A+tan B1−tan A.tan B>0 [∵tan A>0,tan B>0,tan C<0] =tan A.tan B<1⇒q<1

In a triangle ABC, if angle C is obtuse and angles A and B are given by roots of the equation $t a n^{2} x + p t a n x + q = 0$ , then the value of q is

The correct option is B
less than 1

We have $A + B = π - C$

$= t a n (A + B) = - t a n C$

$= \frac{t a n A + t a n B}{1 - t a n A . t a n B} > 0$ $[∵ t a n A > 0, t a n B > 0, t a n C < 0]$

$= t a n A . t a n B < 1 \Rightarrow q < 1$