prove that the left side is equal to the right side without using the right side
tan(t/2) cos²(t) - tan(t/2) = (sin(t)/sec(t)) - sin(t)
=> tan(t/2) cos²(t) - tan(t/2) = sin(t)*cos(t) - sin(t)
=> tan(t/2) [ cos²(t) - 1] = sin(t) [ cos(t) - 1 ]
....................sin(t) [ cos(t) - 1 ]
=> tan(t/2) = ------------------------- ; now use A² - B² = (A + B)*(A - B) at DENOMINATOR
......................[ cos²(t) - 1]
....................sin(t) [cos(t) - 1]
=> tan(t/2) = -------------------------------- ; now cancel out {cos(t) - 1}
....................[cos(t) - 1]*[cos(t) + 1]
.......................sin(t)
=> tan(t/2) = ---------------- ; now use sin 2A = 2sinA*cosA and cos 2A = 2cos²A - 1
....................[cos(t) + 1]
....................2sin(t/2)*cos(t/2)
=> tan(t/2) = ------------------------- ; now cancel out 2cos(t/2) , as -1 + 1 = 0
....................[2cos²(t/2) - 1+ 1]
....................sin(t/2)
=> tan(t/2) = ---------- = tan (t/2)
....................cos(t/2)
=> LHS = RHS Hence proved
Hope this detailed helped you,
Vick
recall identities sin2A = 2sinA.cosA, cos2A = 2cos²A -1
which can be used for finding half angle identities
LHS = tan(T/2).(1-cos²T)
= tan(T/2).(sin²T)
= [sin(T/2)/cos(T/2) ] [4sin²(T/2)/cos²(T/2)]
= [2sin(T/2).cos(T/2)].2sin²(T/2)
= [sinT].[2cos²(T/2)-2]
= [sinT].[(2cos²(T/2)-1) -1]
= sinT(cosT - 1)
= sinT/secT - sinT
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Verified answer
tan(t/2) cos²(t) - tan(t/2) = (sin(t)/sec(t)) - sin(t)
=> tan(t/2) cos²(t) - tan(t/2) = sin(t)*cos(t) - sin(t)
=> tan(t/2) [ cos²(t) - 1] = sin(t) [ cos(t) - 1 ]
....................sin(t) [ cos(t) - 1 ]
=> tan(t/2) = ------------------------- ; now use A² - B² = (A + B)*(A - B) at DENOMINATOR
......................[ cos²(t) - 1]
....................sin(t) [cos(t) - 1]
=> tan(t/2) = -------------------------------- ; now cancel out {cos(t) - 1}
....................[cos(t) - 1]*[cos(t) + 1]
.......................sin(t)
=> tan(t/2) = ---------------- ; now use sin 2A = 2sinA*cosA and cos 2A = 2cos²A - 1
....................[cos(t) + 1]
....................2sin(t/2)*cos(t/2)
=> tan(t/2) = ------------------------- ; now cancel out 2cos(t/2) , as -1 + 1 = 0
....................[2cos²(t/2) - 1+ 1]
....................sin(t/2)
=> tan(t/2) = ---------- = tan (t/2)
....................cos(t/2)
=> LHS = RHS Hence proved
Hope this detailed helped you,
Vick
recall identities sin2A = 2sinA.cosA, cos2A = 2cos²A -1
which can be used for finding half angle identities
LHS = tan(T/2).(1-cos²T)
= tan(T/2).(sin²T)
= [sin(T/2)/cos(T/2) ] [4sin²(T/2)/cos²(T/2)]
= [2sin(T/2).cos(T/2)].2sin²(T/2)
= [sinT].[2cos²(T/2)-2]
= [sinT].[(2cos²(T/2)-1) -1]
= sinT(cosT - 1)
= sinT/secT - sinT
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