Asked by Satyam Singh
prove that tan(pi/4 + 1/2cos-1a/b) + tan(π/4-1/2cos-1a/b)=2b/a
Answers
Answered by
Steve
tan x/2 = (1-cosx)/sinx
so,
tan(1/2 cos^-1(a/b)) = (1-a/b)/(√(b^2-a^2)/b)
= √((b-a)/(b+a))
so, if we let z = 1/2 cos^-1(a/b)
tan(π/4+z)+tan(π/4+z) = (1+√((b-a)/(b+a)))/(1-√((b-a)/(b+a))) + (1-√((b-a)/(b+a)))/(1+√((b-a)/(b+a)))
putting all over a common denominator, things cancel out quite nicely, leaving
= 2b/a
so,
tan(1/2 cos^-1(a/b)) = (1-a/b)/(√(b^2-a^2)/b)
= √((b-a)/(b+a))
so, if we let z = 1/2 cos^-1(a/b)
tan(π/4+z)+tan(π/4+z) = (1+√((b-a)/(b+a)))/(1-√((b-a)/(b+a))) + (1-√((b-a)/(b+a)))/(1+√((b-a)/(b+a)))
putting all over a common denominator, things cancel out quite nicely, leaving
= 2b/a
Answered by
Ritesh
If tanA=x
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