UP Board Important Question
Mathematics (Conditional Trigonometric Identities)
Class – XII
[Set-2]

1. cosA + cosB – cosC = 4 cosA/2 . cosB/2 . sinC/2 – 1

Solution:

L.H.S. = (cosA + cosB) – cosC

          = 2 cos (A + B)/2 . cos (A - B)/2 - cosC since A + B + C = π

          = 2 cos (π - C)/2 . cos (A - B)/2 - cosC

          = 2 cos (π/2 – C/2) . cos (A - B)/2 - cosC

          = 2 sinC/2 . cos (A - B)/2 - cosC since cosC = 1 – 2 sin²C/2

          = 2 sinC/2 . cos (A - B)/2 - 1 + 2 sin²C/2

          = 2 sinC/2 [cos (A - B)/2 + sinC/2] -1

          = 2 sinC/2 [cos (A - B)/2 + sin{π – (A + B)}/2] -1

          = 2 sinC/2 [cos (A - B)/2 + cos (A + B)/2] -1

          = 2 sinC/2 [cos (A + B)/2 + cos (A - B)/2] -1

          = 2 sinC/2 [2 cos{(A + B) + (A - B)}/2x2 . cos{(A + B) - (A - B)}/2x2] -1

          = 4 sinC/2 . cos2A/4 . cos2B/4 – 1

          = 4 sinC/2 . cosA/2 . cosB/2 – 1

          = 4 cosA/2 . cosB/2 . sinC/2 – 1 Proved

2. sinA/2 + sinB/2 + sinc/2 = 1 + 4 sin (π -A)/4 . sin (π -B)/4 . sin (π -C)/4

3. cos²A + cos²B - cos²C = 1 - 2 sinA . sinB . sinC

Solution:

L.H.S. = cos²A + cos²B - cos²C

           = cos²A + (1 - sin²B) - cos²C

           = 1 + (cos²A - sin²B) - cos²C

          = 1 + cos(A + B) . cos(A - B) - cos²C

          = 1 + cos(π - C) . cos(A - B) - cos²C

          = 1 + (- cosC) . cos(A – B) - cos²C

          = 1 – cosC [. cos(A – B) + cosC ]

          = 1 – cosC [cos(A – B) + cos {π – (A + B)} ]

          = 1 – cosC [cos(A – B) - cos(A + B) ]

          = 1 – cosC [2 sin (A – B + A + B)/2 . sin (A + B – A + B)/2]

          = 1 – cosC [2 sin2A/2 . sin2B/2]

          = 1 – cosC [2 sinA . sinB]

          = 1 - 2 sinA . sinB . sinC Proved

4. 1 - 2 sinB . sinC . sinA + cos²A = cos²B + cos²C

5. cos²A + cos²B + cos²C = 1 - 2 cosA . cosB . cosC

6. tanA + tanB + tanC = tanA . tanB . tanC

7. cotB . cotC + cotC . cotA + cotA . cotB = 1

Solution:

Since A + B + C = π => A + B = π – C

                               => cot(A + B) = cot(π - C)

                               => (cotA.cotB – 1)/ cotB + cotA = - cotC

                               => cotA.cotB – 1 = - cotB . cotC – cotA . cotC

                               => cotB . cotC + cotC . cotA + cotA . cotB = 1 Proved

8. tanA/2 . tanB/2 + tanB/2 . tanC/2 + tanC/2 . tanA/2 = 1

9. (cotB + cotC)/(tanB + tanC) + (cotC + cotA)/(tanC + tanA) + (cotA + cotB)/(tanA + tanB) = 1

10. tanA . tanB + tanB . tanC + tanC . tanA = 1

11. tanA/2 + tan(B+C)/2 = secA/2 . sec(B + C)/2

12. sin2A – sin2B = sin(A + B) . sin(A - B)

13. sin2A/(1 + cos2A) = tanA

14. (sinA + sin2A)/(1 + cosA + cos2A) = tanA

15. (sinA + sin3A)/(cosA + cos3A) = tan2A