Question

Statement-1 : In ∆ABC, tanA = tanB = p and tanC = 2 then number of possible values of 'p' is 1.

and

Statement-2 : If tanA + tanB + tanC = tanA. tanB.tanC, then A,B,C are angles of triangle ABC.

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False.
• D. Statement-1 is False, Statement-2 is True.

Answer 1

Answer: (C)

Statement-1 is True, Statement-2 is False.

Answer 2

Detailed analysis of Statement-1:

Statement-1: In $\triangle ABC$, $\tan A = \tan B = p$ and $\tan C = 2$ then number of possible values of 'p' is 1.

For angles $A, B, C$ of a triangle, their sum is $\pi$ radians ($180^\circ$). $A + B + C = \pi$ This implies $\tan(A+B+C) = \tan(\pi) = 0$. The general identity for the tangent of a sum of three angles is:

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

For $\tan(A+B+C) = 0$, the numerator must be zero (provided the denominator is not zero). Thus, for a triangle (assuming no angle is $\pi/2$):

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

Substitute the given values $\tan A = p$, $\tan B = p$, and $\tan C = 2$:

$$p + p + 2 = p \cdot p \cdot 2$$ $$2p + 2 = 2p^2$$

Divide the entire equation by 2:

$$p + 1 = p^2$$

Rearrange into a standard quadratic equation:

$$p^2 - p - 1 = 0$$

Solve for $p$ using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$$ $$p = \frac{1 \pm \sqrt{1 + 4}}{2}$$ $$p = \frac{1 \pm \sqrt{5}}{2}$$

This gives two possible values for $p$:

$p_1 = \frac{1 + \sqrt{5}}{2}$

$p_2 = \frac{1 - \sqrt{5}}{2}$

Now, we must check if these values of $p$ lead to a valid triangle. For $A, B, C$ to be angles of a triangle, they must satisfy $0 < A, B, C < \pi$. Also, at most one angle can be obtuse (i.e., greater than $\pi/2$). Given $\tan C = 2$. Since $2 > 0$, $C$ must be an acute angle ($0 < C < \pi/2$). Specifically, $C = \arctan(2) \approx 63.43^\circ$. Given $\tan A = p$ and $\tan B = p$, this implies $A=B$. If $p$ were negative, then $\tan A < 0$ and $\tan B < 0$. This would mean $A$ and $B$ are both obtuse angles (angles between $\pi/2$ and $\pi$). If $A > \pi/2$ and $B > \pi/2$, then $A+B > \pi$. This would make $A+B+C > \pi$, which is impossible for a triangle (as $C > 0$). Therefore, for a valid triangle, $p$ must be positive.

Let's evaluate the two values of $p$:

$p_1 = \frac{1 + \sqrt{5}}{2} \approx \frac{1 + 2.236}{2} \approx 1.618$. This value is positive.

$p_2 = \frac{1 - \sqrt{5}}{2} \approx \frac{1 - 2.236}{2} \approx -0.618$. This value is negative.

Since $p$ must be positive, only $p = \frac{1 + \sqrt{5}}{2}$ is a valid solution. For this value, $\tan A = \tan B = \frac{1+\sqrt{5}}{2} > 0$, so $A$ and $B$ are acute angles. Since $A, B, C$ are all acute, their sum is less than $3\pi/2$. Since the sum is $\pi$, this forms a valid triangle. Thus, there is only 1 possible value for 'p'. Statement-1 is True.

Detailed analysis of Statement-2:

Statement-2: If $\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$, then $A, B, C$ are angles of triangle ABC.

The given condition is $\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0$. This is the numerator of the expression for $\tan(A+B+C)$. So, the condition implies $\tan(A+B+C) = 0$, provided the denominator $1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)$ is not zero. If $\tan(A+B+C) = 0$, then $A+B+C = n\pi$ for some integer $n$.

For $A, B, C$ to be angles of a triangle, they must satisfy two conditions:

1. $0 < A, B, C < \pi$ (each angle must be positive and less than $180^\circ$).
2. $A+B+C = \pi$ (their sum must be exactly $180^\circ$).

Let's test if the given condition necessarily implies $A+B+C = \pi$. Consider a counterexample: Let $A=B=C=120^\circ$. In this case, $A+B+C = 120^\circ + 120^\circ + 120^\circ = 360^\circ = 2\pi$. These angles do not form a triangle because their sum is $2\pi$, not $\pi$. Now, let's check if they satisfy the given condition: $\tan A = \tan(120^\circ) = -\sqrt{3}$ $\tan B = \tan(120^\circ) = -\sqrt{3}$ $\tan C = \tan(120^\circ) = -\sqrt{3}$ Left Hand Side (LHS) of the condition: $\tan A + \tan B + \tan C = (-\sqrt{3}) + (-\sqrt{3}) + (-\sqrt{3}) = -3\sqrt{3}$ Right Hand Side (RHS) of the condition: $\tan A \cdot \tan B \cdot \tan C = (-\sqrt{3}) \cdot (-\sqrt{3}) \cdot (-\sqrt{3}) = (3) \cdot (-\sqrt{3}) = -3\sqrt{3}$ Since LHS = RHS ($-3\sqrt{3} = -3\sqrt{3}$), the condition $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ holds for $A=B=C=120^\circ$. However, as shown, these angles do not form a triangle. Therefore, the statement "If $\tan A + \tan B + \tan C = \tan A \tan B \tan C$, then $A, B, C$ are angles of triangle ABC" is False.

Conclusion: Statement-1 is True. Statement-2 is False.

This corresponds to option (C).