How do I simplify tan θ + cot θ ?
Or is it already simplified?
you could do ....
sinx/cosx + cosx/sinx
= (sin^2x + cos^2x)/(sinxcosx)
= 1/[ (1/2)2sinxcosx ]
= 1/((1/2)sin 2x)
= 2/sin2x or 2csc 2x
To simplify tan θ + cot θ, we can use the reciprocal identities for tangent and cotangent.
Recall that:
1. tan θ = sin θ / cos θ
2. cot θ = 1 / tan θ = cos θ / sin θ
Now let's simplify:
tan θ + cot θ = (sin θ / cos θ) + (cos θ / sin θ)
To add these fractions, we need a common denominator. The common denominator is cos θ * sin θ:
= (sin^2 θ + cos^2 θ) / (cos θ * sin θ)
Now, since sin^2 θ + cos^2 θ = 1 (from the Pythagorean identity), we can simplify further:
= 1 / (cos θ * sin θ)
So, tan θ + cot θ simplifies to 1 / (cos θ * sin θ).
Therefore, the expression is not already simplified.
To simplify the expression tan θ + cot θ, we can use the reciprocal identities of tangent and cotangent. The reciprocal of tangent is cotangent, and the reciprocal of cotangent is tangent.
Starting with tan θ + cot θ, we can rewrite cot θ as 1/tan θ using the reciprocal identity of cotangent. Now, the expression becomes tan θ + 1/tan θ.
To combine these fractions, we need a common denominator. The common denominator here is tan θ. So, we multiply the second term, 1/tan θ, by tan θ/tan θ. This gives us:
tan θ + 1/tan θ = (tan^2 θ + 1) / tan θ.
So, the expression tan θ + cot θ can be simplified to (tan^2 θ + 1) / tan θ.
This expression cannot be simplified further unless you have other information about θ or if there are specific instructions provided in the problem you are working on. Therefore, the expression (tan^2 θ + 1) / tan θ is the simplified form of tan θ + cot θ.