 # Q.1 tanx + tan 2x + tan 3x = 0Q.2 sec x - 1 = ( 1.414 - 1) tanxAns. 1.   n3.14/3, [ n3.14  (+-)  tan inverse 1/1.414]ans 2.   2n3.14 + 3.14/4

12 years ago

Dear student

tanx+tan2x+tan3x=0

(tanx+tan2x)(1-tanxtan2x)/((1-tanxtan2x)+tan3x=0

tan3x(1-tanxtan2x)+tan3x=0

so we get

either tan3x=0 ortanxtan2x=2

Solve for x....

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Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

4 years ago

tanx + tan2x+ tan3x=0
or, tanx+ tan2x= - tan3x
or, (sinx/cosx)+(sin 2x /Cos 2x)=-(sin 3x /cos 3x)
or, (sin x *cos 2 X + cos x *sin 2x)/cos x * cos 2x = -sin 3x / cos 3x
or, sin( 2x+x)* cos 3x= - cos x *cos 2x *sin 3x
or, sin 3x*cos 3x+ cos x * cos 2x *sin 3x =0
or, sin 3x ( cos 3x + cos x *cos 2x)=0
or, sin 3x (cos ( 2x + x)+ cos x * cos 2x)=0
or, sin 3x (cos x *cos 2x - sin x *sin 2x+ cos x *cos 2x) =0
or, -sin 3x *sin x * sin 2x=0

Either sin3x=0
i.e, 3x=nπ
i.e, X=nπ/3

or, sin 2x=0
i.e, 2x=nπ
i.e, X= nπ/2

or, sin x= 0
i.e, X=nπ
Required solution is X= nπ/3