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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If A > 0, B > 0 and A + B = $\frac{\pi}{6}$ , then the minimum value of $tanA + tanB$
Solution
On differentiatingx= tanA + tan(π/6-A)
we get :
dx/dA = sec²A-sec²(π/6-A)
now putting
dx/dA=0
we get
cos²(A) = cos²(π/6-A) so 0≤A≤π/6
therefore
A=π/6-A from here we get A = π/12 = B
so minimum value of that function is
2tanπ/12 which is equal to 2(2-√3)
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