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Grade: 11

                        

if alpha and beta aare the solution of the equation: atan(theta)+bsec(theta)=c find tan(alpha+beta)

6 years ago

Answers : (4)

Rinkoo Gupta
askIITians Faculty
80 Points
							atan@+bsec@=c
bsec@=c-atan@
squaring bothsides
b^2sec^2@=(c-atan@)^2
b^2(1+tan^2@)=c^2+a^2tan^2@-2actan@
(b^2-a^2)tan^2@+2actan@+b^2-c^2=0
since alpha and beta are the roots of the eq so
tan(alpha)+tan(beta)=-2ac/(b^2-a^2)
tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)
tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}
=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]
=-2ac/(c^2-a^2)
=2ac/(a^2-c^2) Ans.

Thanks & Regards
Rinkoo Gupta
AskIITians faculty
6 years ago
Jatin
26 Points
							atan@+bsec@=cbsec@=c-atan@squaring bothsidesb^2sec^2@=(c-atan@)^2b^2(1+tan^2@)=c^2+a^2tan^2@-2actan@(b^2-a^2)tan^2@+2actan@+b^2-c^2=0since alpha and beta are the roots of the eq sotan(alpha)+tan(beta)=-2ac/(b^2-a^2)tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]=-2ac/(c^2-a^2)=2ac/(a^2-c^2) Ans.Thanks & Regards
						
3 years ago
Vnk
11 Points
							Sir alpha and beta are roots so how did u take tan alpha and tan beta as roots...Also  how do u know that we have to square??
						
3 years ago
ShAdOw
11 Points
							Hiksksjsjjsb bsbjsjsjbs. Shahajahan s aajjajajsb s sheuuw8wiwpqdbd s sjdjbskaoqbw. Susiwkeb a aowjebd. Alaowiwhge d sa lqowiebd. Amqlwoeiry829wnd d shakhs wi2ia s ek2o3hbs qkwieva. Qiwieheuowdbd qkiwjwna sbwiowbe
						
2 years ago
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