tan1/2=cscø-cotø
I will assume you are solving for ø
cscø-cotø = 1/2
1/sinø - cosø/sinø = 1/2
(1-cosø)/sin = 1/2
1 - cosø = sinø/2
square both sides
1 - 2cosø + cos^2 ø = sin^2 ø/4
4 - 8cosø + 4cos^2 ø = 1 - cos^2 ø
5cos^2 ø - 8cosø + 3 = 0
(cosø - 1)(5cosø - 3) = 0
cosø = 1 or cosø = 3/5
ø = 0, 360 or ø = appr 53.13° , 306.87°
but if ø = 0 or 360° , both cscø and cotø are undefined
furthermore, since I squared both sides, all answers must
be verified. ø=53.13 works, but ø = 306.87 gives me -1/2, not 1/2
so for 0 < ø < 360° , ø = 53.13°
Or, if you are trying to prove that
tan1/2 ø=cscø-cotø
start with
cos2ø = 2cos^2ø-1 = 1-2sin^2ø
Then
tan ø/2 = sin(ø/2)/cos(ø/2)
= √((1-cosø)/2) / √((1+cosø)/2)
= √(1-cosø)/√(1+cosø)
= √(1-cosø)^2 /√((1+cosø)(1-cosø))
= (1-cosø) / √(1-cos^2ø)
= (1-cosø) / √sin^2ø
= (1-cosø)/sinø
= cscø-cotø
To prove the equation tan(1/2) = cscø - cotø, we can start by using the definitions of the trigonometric functions.
1. Start with the left-hand side (LHS) of the equation:
tan(1/2) = sin(1/2) / cos(1/2)
2. Recall the half-angle identities:
sin(θ/2) = ±√[(1 - cosθ)/2]
cos(θ/2) = ±√[(1 + cosθ)/2]
tan(θ/2) = sinθ / (1 + cosθ)
3. Apply the half-angle identity for sine to sin(1/2):
sin(1/2) = ±√[(1 - cos1)/2]
4. Since we're dealing with the angle 1/2, we can let ø = 1.
Therefore, cos1 = cosø.
5. Substitute cos1 with cosø in the previous step:
sin(1/2) = ±√[(1 - cosø)/2]
6. Now, let's move to the right-hand side (RHS) of the equation:
cscø - cotø = 1/sinø - cosø/sinø
7. Simplify the RHS by finding a common denominator:
cscø - cotø = (1 - cosø) / sinø
8. Notice that the RHS is also equal to the simplified form of sin(1/2):
(1 - cosø) / sinø = ±√[(1 - cosø)/2]
9. Since both sides of the equation are equal to ±√[(1 - cosø)/2], we can conclude that:
tan(1/2) = cscø - cotø
Therefore, the equation tan(1/2) = cscø - cotø is proven.