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Ask iit jee aieee pet cbse icse state board experts Expert Question: re:trigo doubt
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[Post New]19 Aug 2007 21:38:44 IST
    Subject: re:trigo doubt
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rohitkuruvila (311)

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sorry ,i mistyped the question earlier.the actual question is:
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
    
[Post New]19 Aug 2007 21:58:27 IST
    Subject: Re:re:trigo doubt
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123santom (14)

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OPTIONS
 
1
 
3 ARE CORRECT!!!
 
 
 
CHEERS!!!!!!!!
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[Post New]19 Aug 2007 22:15:00 IST
    Subject: Re:re:trigo doubt
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aankurverma (1310)

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cos(A - B) = 3/5
cosAcosB + sinAsinB = 3/5
divide by cosAcosB
cosAcosB = 3/5*1/(1 + tanAtanB)
cosAcosB = 1/5
sinAsinB = 3/5 - 1/5 = 2/5
cos(A + B) =  1/5 - 2/5 = -1/5
 
1 n 3 r correct hope u got it do rate me if helpfull
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[Post New]19 Aug 2007 22:55:18 IST
    Subject: Re:re:trigo doubt
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iitkgp_bipin (6144)

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tanA.tanB = 2

(sinA.sinB) / (cosA.cosB) = 2

sinA.sinB = 2cosA.cosB

Now cos(A-B)=3/5

cosA.cosB + sinA.sinB = 3/5

cosA.cosB + 2cosA.cosB = 3/5

cosA.cosB = 1/5............(1)

so sinA.sinB = 2cosA.cosB = 2/5.............(2)

cos(A+B) = cosA.cosB - sinA.sinB = 1/5 - 2/5 = -1/5...........(3)

so third option is correct.

 
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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