Question

With usual notations in $\triangle ABC$, if $\angle B = \pi/2$, and $\tan A, \tan C$ are roots of equation $px^2 + qx + r = 0, p \neq 0$, then ______.

• A. $p + q = r$
• B. $r + p = q$
• C. $r = p$
• D. $p = q$

Hint:

In a right-angled triangle, the acute angles are always complementary. This immediately implies that the product of their tangents is exactly 1 ($\tan \theta \cdot \tan(90^\circ-\theta) = 1$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a right-angled triangle at B, meaning $\angle B = 90^\circ$.
We are also told that the tangents of the other two angles, $\tan A$ and $\tan C$, are the roots of the given quadratic equation. We need to find the relation between the coefficients $p, q$, and $r$.

Step 2: Key Formula or Approach:
For any quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:
Product of roots ($\alpha \cdot \beta$) = $\frac{c}{a}$.
In a right-angled triangle ($\angle B = \pi/2$), the sum of the remaining two angles is $\angle A + \angle C = \pi/2$.

Step 3: Detailed Explanation:
Since $A + C = 90^\circ$, we can write $C = 90^\circ - A$.
Applying the tangent function:
$$\tan C = \tan(90^\circ - A) = \cot A$$
The roots of the equation $px^2 + qx + r = 0$ are given as $\tan A$ and $\tan C$.
Let's find the product of these roots:
$$\text{Product of roots} = \tan A \cdot \tan C$$
Substitute $\tan C$ with $\cot A$:
$$\text{Product of roots} = \tan A \cdot \cot A = \tan A \cdot \frac{1}{\tan A} = 1$$
From the quadratic equation $px^2 + qx + r = 0$, the product of roots is theoretically $\frac{r}{p}$.
Equating the two values:
$$\frac{r}{p} = 1$$
$$r = p$$

Step 4: Final Answer:
The required relation is $r = p$, which corresponds to option (c).