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rachio
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#1
Report Thread starter 13 years ago
#1
The curve C has the equation y = tanx + lnsecx

At the point P on C, whose x-coordinate is p, the gradient is 3

Show the tanp = -2


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yazan_l
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#2
Report 13 years ago
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y = tanx + lnsecx
dy/dx = sec²x + tanx
3 = sec²p + tanp
3 = (tan²p + 1) + tanp
tan²p + tanp - 2 = 0
( tanp + 2 ) ( tanp - 1 ) = 0
( tanp + 2 ) = 0
tanp = -2

or tanp = 1
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rachio
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#3
Report Thread starter 13 years ago
#3
Thank you very much - it seems so easy when someone else does it!!!
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