sec4(Θ) - sec2(Θ) = tan4(Θ) + tan2(Θ)
Factor each side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) [tan2(Θ) + 1]
Use the identitiy: 1 + tan2(Θ) = sec2(Θ)
We can also write it as: tan2(Θ) = sec2(Θ) - 1
Substitute the first form of the identity in the right side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) sec2(Θ)
Substitute the second form of the identity in the left side: sec2(Θ) tan2(Θ) = tan2(Θ) sec2(Θ)
Is that close enough for you ?
Is it worth a trust point ?
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
Yes. (Theta in radians, and then approximately, not exactly.)
theta = arcsin(0.0138) is the principal value.
It also equals 13 12.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
Yes. (Theta in radians, and then approximately, not exactly.)
theta = arcsin(0.0138) is the principal value.
Yes, it is.
It also equals 13 12.
Theta equals 0 or pi.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
No.
The answer depends on what theta is and the units of its measurement.
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
If r-squared = theta then r = ±sqrt(theta)