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Tan-ing the sine (Posted on 2019-03-03) Difficulty: 3 of 5
tan2a.tan2b + tan2b.tan2c + tan2c.tan2a + 2tan2a.tan2b.tan2c = 1

If a, b, c satisfy the equation above, find the value of sin2a + sin2b + sin2c.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution | Comment 1 of 3
Call D the above expression  D=1

tan
2a.tan2b=(1-cos2a)*(1-cos2b)/cos2a.cos2b
tan2b*tan2c= ...

Developping

D=1-[1-(sin2a+sin2b+sin2c)]/cos2a*cos2b*cos2c)=1

Then: 

1-(sin2a+sin2b+sin2c)=0 

and

sin2a+sin2b+sin2c=1

Edited on March 5, 2019, 7:10 am
  Posted by armando on 2019-03-05 06:55:11

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