Question #09530

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Answer 1 · 2016-12-31T09:21:40

To prove
$2tan^-1(tan (A/2) * tan(pi/4 - B/2))=tan^-1((sinAcosB)/(cosA+sinB))$

Let $A/2=x and (pi/4-B/2)=y$
so
$2x=Aand 2y=pi/2-B$

Now

$LHS=2tan^-1(tan (A/2) * tan(pi/4 - B/2))$

$=2tan^-1(tanx * tany)$

$=tan^-1((2tanx * tany)/(1-tan^2x * tan^2y))$

$=tan^-1((2tanx * tany*cos^2x*cos^2y)/(cos^2xcos^2y-tan^2xtan^2y*cos^2xcos^2y))$

$=tan^-1((2sinx cosx*2sinycosy)/(2(cos^2xcos^2y-sin^2xsin^2y)))$

$=tan^-1((sin2x sin2y)/(2(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)))$

$=tan^-1((sin2x sin2y)/(2cos(x+y)cos(x-y)))$

$=tan^-1((sin2x sin2y)/(cos(2x)+cos(2y)))$

$=tan^-1((sinA sin(pi/2-B))/(cosA+cos(pi/2-B)))$

$=tan^-1((sinA cosB)/(cosA+sinB))=RHS$

Proved