perplexus.info :: Just Math : Tan-ing the sine

Tan-ing the sine (Posted on 2019-03-03)

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

No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)

Solution

| Comment 1 of 3

Call D the above expression D=1

tan²a.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

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info