Prove that 1+cosx+sinx/1-cosx+sinx=cotx
2 answers
I do not know what are your numerators and denominators.
multiply top and bottom by 1+sin+cos
For ease of reading, I'll just use s and c for sinx and cosx
(1+s+c)/(1+s-c) * (1+s+c)/(1+s+c)
= (1+s+c)^2 / ((1+s)^2 - c^2)
= (1+s^2+c^2+2s+2c+2sc) / (1+2s+s^2-c^2)
= (2+2s+2c+2sc)/(1+2s+s^2-1+s^2)
= (2+2s+2c+2sc)/(2s+2s^2)
= ((1+c)(1+s))/(s(1+s))
= (1+c)/s
Now you see the typo. Using the half-angle formulas, the result is
cot(x/2), not cot(x)
For ease of reading, I'll just use s and c for sinx and cosx
(1+s+c)/(1+s-c) * (1+s+c)/(1+s+c)
= (1+s+c)^2 / ((1+s)^2 - c^2)
= (1+s^2+c^2+2s+2c+2sc) / (1+2s+s^2-c^2)
= (2+2s+2c+2sc)/(1+2s+s^2-1+s^2)
= (2+2s+2c+2sc)/(2s+2s^2)
= ((1+c)(1+s))/(s(1+s))
= (1+c)/s
Now you see the typo. Using the half-angle formulas, the result is
cot(x/2), not cot(x)
0 answers