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Trignometry

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31 Dec 2011 12:59:33 IST

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solution of triangle

Mathematics

if in a triangle angle C =90 degreees then find the value of (a^2+b^2)/(a^2-b^2) X sin (A-B)

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pritam sarma

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Joined: 2 Jan 2012

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2 Jan 2012 09:02:31 IST

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if in a triangle c=90 degrees, find (a^2+b^2)/a^2-b^2) X sin(A-B)

ramreddysrinivas

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3 Jan 2012 15:31:36 IST

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put a =2RsinA

b= 2RsinB

write sin²-sin²B=sin(A+B)sin(A-B)

then simplify

the answer is 1.

ramreddysrinivas

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3 Jan 2012 15:32:59 IST

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put a =2RsinA

b= 2RsinB

write sin²A-sin²B=sin(A+B)sin(A-B)

then simplify

the answer is 1.

apoorv

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Joined: 12 Sep 2011

Posts: 64

12 Jan 2012 06:07:17 IST

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a^2+b^2/a^2-b^2 X sin(A-B)

=sin^2 A+sin^2 B/sin^2 A-sin^2 B X sin(A-B)

=sin^2 A+sin^2 B/sin(A+B)

=sin^2 (pi/2-B)+sin^2 B/sin (pi/2) [becoz C=pi/2]

=cos^2 B+sin^2 B/1

=1/1

=1...

plss press like if u really liked my answer!!

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