Isn't sin^2C=1-cos^2C= (1+cosC)(1-cosC) ?
If that is so, then it becomes
1+cosC-(1-cosC)=1
which solves very quickly.
Given that sin^2ceta/1-cos ceta -sin^2ceta/1+cosceta=1 where 0degress is less than or equal to ceta and less than or equal to 360 degress. find the value of ceta
2 answers
write the two terms with a common denomimator. I will call "ceta" x,
sin^2 x [1/(1-cosx) - 1/(1+cos x)]
= sin^2x [(1 + cosx -1 +cosx)/(1-cos^2x)]
= 2 cos x = 1
The solutions are where cos x = 1/2
That would be at x = 60 degrees and 300 degrees.
sin^2 x [1/(1-cosx) - 1/(1+cos x)]
= sin^2x [(1 + cosx -1 +cosx)/(1-cos^2x)]
= 2 cos x = 1
The solutions are where cos x = 1/2
That would be at x = 60 degrees and 300 degrees.