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Solve for t: 2t(sin(t2+1) + cos(t2+1))2 - 4tsin(t2+1)cos(t2+1) = 2.
2t[sin(t^2+1) + cos(t^2+1)]^2 - 4tsin(t^2+1)cos(t^2+1) = 2. ----(1)
Put (t^2+1) = y
2t(siny + cosy)^2 - 4tsinycosy = 2.
t[(siny + cosy)^2 - 2sinycosy] = 2.
t[(sin^2y+cos^2y +2sinycosy)- 2sinycosy] = 12
(using (a+b)^2 = a^2+b^2+2ab where here a =suny and b = cosy )
t[(sin^2y+cos^2y) +(2sinycosy- 2sinycosy)] = 2 ----(*)
(by formula sin^2y+cos^2y = 1 )
t[(1)+0] = 2
tX(1) = 2
That is t = 2
Answer: t = 2
Remark: If you are wondering about the quantity y or for that matter (t^2+1), the terms in that quantity being of equal in magnitude and opposite in sign have got cancelled by the additive inverse law! as you can see in step (*)