Question

Match the items of List-I with those of List-II (Here \( \Delta \) denotes the area of \( \triangle ABC \)).

Then the correct match is

• A. A-VI; B-I; C-II; D-III
• B. A-II; B-I; C-V; D-III
• C. A-II; B-VI; C-V; D-I
• D. A-VI; B-II; C-I; D-IV

Hint:

For matching-type problems in trigonometry and geometry, express each term in known formulas and find corresponding values systematically.

Answer 1

The Correct Option is B

A. ΣcotA

This represents the sum of the cotangents of the angles of a triangle. It is equivalent to (a² + b² + c²) / 4Δ where a, b, and c are the side lengths and Δ is the triangle's area. 
Therefore, A matches with II.

B. Σcot(A/2)

This represents the sum of the cotangents of the half-angles. This equals (a + b + c)² / 4Δ. 
Thus, B matches with I.

C. tanA : tanB : tanC = 1:2:3

The ratio of sines of the angles is proportional to the sides opposite to the respective angles. So, sin A : sin B : sin C = √5 : 2√2 : 3. 
C matches with V.

D. cot(A/2) : cot(B/2) : cot(C/2) = 3:7:9

The ratio of cotangents of half-angles is proportional to (s-a):(s-b):(s-c), where 's' is the semi-perimeter. Since a+b+c = 2s, the sides a:b:c are proportional to 12:5:13. 
Therefore, D matches with III.

Answer 2

A. \(\Sigma \cot A\)

The sum of cotangents of the angles of a triangle can be found by expressing \(\cot A\) as \(\frac{\cos A}{\sin A}\). Using the Law of Cosines, write each cosine in terms of the sides:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] and similarly for \(\cos B\) and \(\cos C\). Using the Law of Sines, \(\sin A = \frac{2\Delta}{bc}\). Substituting these into \(\cot A = \frac{\cos A}{\sin A}\) and simplifying for all angles yields:
\[ \Sigma \cot A = \frac{a^2 + b^2 + c^2}{4\Delta} \] Thus, A corresponds to II.

B. \(\Sigma \cot \frac{A}{2}\)

Recall the half-angle cotangent relation:
\[ \cot \frac{A}{2} = \frac{s - a}{r} \] where \(r\) is the inradius and \(s\) is the semi-perimeter. Summing over all angles gives:
\[ \Sigma \cot \frac{A}{2} = \frac{(s - a) + (s - b) + (s - c)}{r} = \frac{3s - (a+b+c)}{r} = \frac{s}{r} \] Since \(\Delta = r s\), this can be rearranged as:
\[ \Sigma \cot \frac{A}{2} = \frac{(a + b + c)^2}{4\Delta} \] Hence, B corresponds to I.

C. \(\tan A : \tan B : \tan C = 1 : 2 : 3\)

Given the ratio of tangents, rewrite it as:
\[ \frac{\sin A}{\cos A} : \frac{\sin B}{\cos B} : \frac{\sin C}{\cos C} = 1 : 2 : 3 \] Assuming a proportionality for the cosines, the sines are in proportion. Using the Law of Sines:
\[ a : b : c = \sin A : \sin B : \sin C = \sqrt{5} : 2\sqrt{2} : 3 \] This shows C corresponds to V.

D. \(\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9\)

Cotangents of half-angles are proportional to:
\[ (s - a) : (s - b) : (s - c) = 3 : 7 : 9 \] With \(a + b + c = 2s\), set:
\[ a = s - 3k, \quad b = s - 7k, \quad c = s - 9k \] Solving for side ratios yields:
\[ a : b : c = 12 : 5 : 13 \] Therefore, D corresponds to III.