Assertion (A) : If x+y+z=xyz then ∑2x1−x2=π2x1−x2 Reason (R): If tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C then A+B+C=nπ,n∈N

Let x=tan⁡A,y=tan⁡B,z=tan⁡C,x+y+z=xyz⇒ tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C⇒ A+B+C=nπ or 2A+2B+2C=2nπ or tan⁡2A+tan⁡2B+tan⁡2C=tan⁡2Atan⁡2Btan⁡2C or 2tan⁡A1−tan2⁡A+2tan⁡B1−tan2⁡B+2tan⁡C1−tan2⁡C =2tan2⁡A1−tan2⁡A2tan⁡B1−tan2⁡B2tan⁡C1−tan2⁡C⇒ 2x1−x2+2y1−y2+2z1−z2=2x1−x22y1−y22z1−z2A is true, R is true and R is correct explanation of A

Assertion (A) : If x+y+z=xyz then ∑2x1−x2=π2x1−x2 Reason (R): If tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C then A+B+C=nπ,n∈N

Let x=tan⁡A,y=tan⁡B,z=tan⁡C,x+y+z=xyz⇒ tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C⇒ A+B+C=nπ or 2A+2B+2C=2nπ or tan⁡2A+tan⁡2B+tan⁡2C=tan⁡2Atan⁡2Btan⁡2C or 2tan⁡A1−tan2⁡A+2tan⁡B1−tan2⁡B+2tan⁡C1−tan2⁡C =2tan2⁡A1−tan2⁡A2tan⁡B1−tan2⁡B2tan⁡C1−tan2⁡C⇒ 2x1−x2+2y1−y2+2z1−z2=2x1−x22y1−y22z1−z2A is true, R is true and R is correct explanation of A

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$Assertion (A) : If x + y + z = x y z then \sum (\frac{2 x}{1 - x^{2}}) = π (\frac{2 x}{1 - x^{2}}) Reason (R): If \tan A + \tan B + \tan C = \tan A \tan B \tan C then A + B + C = n π, n \in N$

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$A is true, R is false$

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$A is true, R is true and R is correct explanation of A$

$A is true, R is true and R is not correct explanation of A$

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Detailed Solution

$\begin{array}{ll} Let x = \tan A, y = \tan B, z = \tan C, x + y + z = x y z \\ \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C \\ \Rightarrow A + B + C = n π \\ or 2 A + 2 B + 2 C = 2 n π \\ or \tan 2 A + \tan 2 B + \tan 2 C = \tan 2 A \tan 2 B \tan 2 C \\ or \frac{2 \tan A}{1 - \tan^{2} A} + \frac{2 \tan B}{1 - \tan^{2} B} + \frac{2 \tan C}{1 - \tan^{2} C} \\ = \frac{2 \tan^{2} A}{1 - \tan^{2} A} \frac{2 \tan B}{1 - \tan^{2} B} \frac{2 \tan C}{1 - \tan^{2} C} \\ \Rightarrow \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \frac{2 y}{1 - y^{2}} \frac{2 z}{1 - z^{2}} \end{array}$