IIT JEE , AIEEE Forum - Mathematics , Trignometry

rohitkuruvila

If cos(A+B)=3/5 and tan A.tan B=2,then

(a)cos A.cos B=1/5

(b)sin A.sin B=-2/5

(c)cos(A+B)=-1/5

(d)sin A.cos B=4/5

pink_ele

sinA sinB/(cosAcosB)=2

2cosA cosB =sinA sinB

cos A+B=3/5

-COS A COS B=3/5

SINA SINB=-6/5

SIN(A+B)=4/5

SINA COS B=4/5-COS A SIN B=4/5(NOT POSSIBLE)

NUDGE ME

krishna.gopal

cos(A+B) = cosAcosB-sinAsinB=3/5

dividing by cosAcosB on both sides
1-tanAtanB=1-2 =(3/5)/(cosAcosB)
So CosACosB=-3/5
SinASinB = CosACosB-3/5= -6/5
Which is not possible
Some thing is wrong with this question

rohitkuruvila

no..I am sure that this question is right

aratrika

cos (A+B) =3/5

cos A cosB - sinA sinB= 3/5 (1`)

GIV: tanAtan B = 2

sin A/cosA * sin B /cosB = 2

or, sin A sin B = 2cosA cosB

or, cosA cos B = (sin A sin B)/2

putting the above value in eqn. (1), we get,

sinAsin B - 2 sinA sin B = 6/5

or, sinAsinB(1 -2) = 6/5

or, sinA sinB = - 6/5

so none of the options are correct .

PLZZZZZZ RATE ME , because i know i'm correct !!!!!

raltz

i think the question shld be cos(A-B) = 3/5

only then we get option (a)

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