(tanθ + cotθ)² = sec²θ + csc²θ
<u>Expand left side</u>: tan²θ + 2tanθcotθ + cot²θ
<u>Evaluate middle term</u>: 2tanθcotθ = = 2
⇒ tan²θ + 2+ cot²θ
= tan²θ + 1 + 1 + cot²θ
<u>Apply trig identity:</u> tan²θ + 1 = sec²θ
⇒ sec²θ + 1 + cot²θ
<u>Apply trig identity:</u> 1 + cot²θ = csc²θ
⇒ sec²θ + csc²θ
Left side equals Right side so equation is verified
Answer:
Step-by-step explanation:
The question presumes you have access to a computer algebra system. The one I have access to provided the output in the attachment. The list at the bottom is the list of the first four derivatives of f(x).
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The derivatives alternate signs, so (-1)^k will be a factor.
The numerators start at 17 and increase by increasing factors: 2, 3, 4, indicating k! will be a factor.
The denominators have a degree that is k+1.
Putting these observations together, we can write an expression for the k-th derivative of f(x):
