explain how to do these with please. I was out a couple of days so give clear instructions.
sin(π/2-ϴ)secϴ
cot(π/2-x)cosϴ
That is indeed hard, someone help this man.
correction cot(π/2-ϴ)cosϴ
To evaluate the expressions sin(π/2-ϴ)secϴ and cot(π/2-x)cosϴ, we'll break it down step by step.
Let's start with the first expression: sin(π/2-ϴ)secϴ.
Step 1: Evaluate sin(π/2-ϴ)
To find sin(π/2-ϴ), we can use the trigonometric identity that sin(π/2-ϴ) = cos(ϴ). So, sin(π/2-ϴ) simplifies to cos(ϴ).
Step 2: Evaluate secϴ
To evaluate secϴ, we need to find the reciprocal of cosϴ. Therefore, secϴ = 1/cosϴ.
Step 3: Substitute the values
Now that we have the simplified versions of sin(π/2-ϴ) and secϴ, we can substitute them back into the original expression.
sin(π/2-ϴ)secϴ = cos(ϴ) * (1/cosϴ)
Simplifying further, we find that cosϴ cancels out:
cos(ϴ) * (1/cosϴ) = 1
So, the expression sin(π/2-ϴ)secϴ simplifies to just 1.
Moving on to the second expression: cot(π/2 - x)cosϴ.
Step 1: Evaluate cot(π/2 - x)
To find cot(π/2 - x), we can use the trigonometric identity that cot(π/2 - x) = tan(x). So, cot(π/2 - x) simplifies to tan(x).
Step 2: Evaluate cosϴ
We already have the value of cosϴ from the previous question, so we can reuse it here.
Step 3: Substitute the values
Now that we have the simplified versions of cot(π/2 - x) and cosϴ, we can substitute them back into the original expression.
cot(π/2 - x)cosϴ = tan(x) * cosϴ
This is the final simplified form of the expression cot(π/2 - x)cosϴ.
In summary, sin(π/2-ϴ)secϴ simplifies to 1, while cot(π/2 - x)cosϴ simplifies to tan(x) * cosϴ.