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Trignometry

Blazing goIITian

Joined:	11 Jun 2009
Post:	386

17 Jun 2009 21:16:10 IST

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369

plz. solve this

None

if [tan(a+b-c)]/[tan(a-b+c)]=[tanc/tanb]

pt sin2a+sin2b+sin2c=0

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®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

23 Jun 2009 17:34:35 IST

Like 1 people liked this

OK can you explain this ?

a=b=c=racpi{4}

rac{ an(a+b-c)}{ an(a-b+c)}=rac{ anleft(rac{pi}{4}ight)}{ anleft(rac{pi}{4}ight)}=1

And,

rac{ an c}{ an b} = 1

But, sin2a+sin2b+sin2c=1+1+1=3

Blazing goIITian

Joined: 11 Jun 2009

Posts: 386

24 Jun 2009 12:01:49 IST

Like 0 people liked this

yep ithink the q is wrong

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