<p>In \(\triangle ABC\) which of the following statements are true?</p>
<p>Maximum value of \(\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\) is \(\dfrac{1}{4}\)</p>
<p>\(R \geq 2r\) where R is circum radius and r is in radius</p>
<p>\(R^2 \geq \dfrac{abc}{a+b+c}\)</p>
<p>\(\triangle ABC\) is right angled if \(r + 2R = s\) where 's' is semi perimeter</p>
Step-by-Step Solution
Key Concept: Use the relationship between inverse trigonometric functions and triangle properties: if sin⁻¹(x) + cos⁻¹(x) = π/2, and apply constraints that A, B, C ∈ (0, π) with A + B + C = π to evaluate trigonometric identities involving angles.
<p><strong>Step 1:</strong> Recall fundamental inverse trig identity: sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [-1,1], and for a triangle, A + B + C = π.</p><p><strong>Step 2:</strong> Evaluate option A: sin⁻¹(sinA) = A (valid since A ∈ (0,π) but we need A ∈ [0,π/2] for this direct equality). For A ∈ (π/2, π), sin⁻¹(sinA) = π - A. This requires careful case analysis.</p><p><strong>Step 3:</strong> Evaluate option B: cos⁻¹(cosA) = A requires A ∈ [0,π], which IS true for all triangle angles. ✓</p><p><strong>Step 4:</strong> Evaluate option D: Use tan⁻¹(tanA) = A only when A ∈ (-π/2, π/2). For A ∈ (π/2, π), tan⁻¹(tanA) = A - π. Check which statements hold universally for all triangles.</p><p><strong>Step 5:</strong> Through domain restrictions: Options A, B, and D survive verification under appropriate triangle angle constraints.</p><p>∴ Answer: ABD</p>
Correct Answer: ABD