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if a and b are the solutions of the equation atan@+bsec@=c then show that tan(a+b)=2ac/(a^2-c^2)
bsec@=c-atan@squaring both sidesb2sec2@=c2+a2tan2@-2actan@b2+b2tan2@=c2+a2tan2@-2actan@(a2-b2)tan2@-2actan@+c2-b2=0since tana & tanb are the roots of the equationtana+tanb=2ac/a2-b2, tanatanb=c2-b2/a2-b2tan(a+b)=tana+tanb/1-tanatab=2ac/a2-b2Thanks & RagardsSunil RaikwaraskIITians Faculty
bsec@=c-atan@squaring both sidesb2sec2@=c2+a2tan2@-2actan@b2+b2tan2@=c2+a2tan2@-2actan@(a2-b2)tan2@-2actan@+c2-b2=0since tana & tanb are the roots of the equationtana+tanb=2ac/a2-b2, tanatanb=c2-b2/a2-b2tan(a+b)=tana+tanb/1-tanatab=2ac/a2-b2Thanks & RegardsRinkoo GuptaAskIITians Faculty
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