Question:

fA+B+C=, then tan⁡(A2)tan⁡(B2)+tan⁡(B2)tan⁡(C2)+tan⁡(C2)tan⁡(A2) is equal to

Updated On: Feb 15, 2024
  • (A) π/6
  • (B) 3
  • (C) 2
  • (D) 1
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The Correct Option is D

Solution and Explanation

Explanation:
Given that, A+B+C=π∴tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2⇒tan⁡B2(tan⁡A2+tan⁡C2)+tan⁡C2tan⁡A2⇒tan⁡B2(sin⁡A2cos⁡C2+sin⁡C2cos⁡A2cos⁡A2cos⁡C2)+sin⁡C2sin⁡A2cos⁡C2cos⁡A2cos⁡A2cos⁡C2C2sin⁡A2⇒tan⁡(B2){sin⁡(A+C2)}+sin⁡C22⇒sin⁡(B/2)+sin⁡(C/2)sin⁡(A/2)cos⁡(A/2)cos⁡(C/2)⇒cos⁡(A+C)/2+sin⁡(C/2)sin⁡(A/2)cos⁡(A/2)cos⁡(C/2)⇒cos⁡(A/2)cos⁡(C/2)−sin⁡(A/2)sin⁡(C/2)+sin⁡(C/2)sin⁡(A/2)cos⁡(A/2)cos⁡(C/2)=cos⁡(A/2)⋅cos⁡(C/2)cos⁡(A/2)⋅cos⁡(C/2)=1

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