In a ΔABC, if \(tan 2A/2+tan 2B/2+tan 2C/2=k\) , then k is always

We can put values of the angles \(A\), \(B\), and \(C\) to get an idea about the value of \(k\): 1. Put \(A = B = C = 60 \): \[ k = tan 2 30 + tan 2 30 + tan 2 30 = 1 \] 2. Put another set of values \(A = B = 30 \) and \(C = 120 \): \[ k = tan 2 15 + tan 2 15 + tan 2 45 > 1 \] Hence, \(k ≥ 1\).

In a ΔABC, if \(tan 2A/2+tan 2B/2+tan... - Solved

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Question 35

Mathematics Trigonometric Equations Hard

In a ΔABC, if \(\tan ^2\frac{A}{2}+\tan ^2\frac{B}{2}+\tan ^2\frac{C}{2}=k\) , then k is always

(A) \(>1\)

(B) \(\geq 1\)

(C) =2

(D) =1

✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We can put values of the angles \(A\), \(B\), and \(C\) to get an idea about the value of \(k\): 1. Put \(A = B = C = 60^\circ\): \[ k = \tan^2 30^\circ + \tan^2 30^\circ + \tan^2 30^\circ = 1 \] 2. Put another set of values \(A = B = 30^\circ\) and \(C = 120^\circ\): \[ k = \tan^2 15^\circ + \tan^2 15^\circ + \tan^2 45^\circ > 1 \] Hence, \(k \geq 1\).

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