if a and b are the solutions of the equation atan@+bsec@=c then show thattan(a+b)=2ac/(a^2-c^2)

bsec@=c-atan@

bsec@=c-atan@
squaring both sides
b2sec2@=c2+a2tan2@-2actan@
b2+b2tan2@=c2+a2tan2@-2actan@
(a2-b2)tan2@-2actan@+c2-b2=0
since tana & tanb are the roots of the equation
tana+tanb=2ac/a2-b2, tanatanb=c2-b2/a2-b2
tan(a+b)=tana+tanb/1-tanatab=2ac/a2-b2

Thanks & Regards

Rinkoo Gupta


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