Cot(a+b) Cot(a-b) can be expressed in terms of cot as well as tan. In this post, we will learn to find cot(a+b)cot(a-b) formula with proof.
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Let us now find the formula of the product cot(a+b) and cot(a-b).
Cot(a+b) Cot(a-b) Formula in Terms of Cot
Using the identities of cot(a+b) and cot(a-b), the given expression can be written as follows:
cot(a+b) cot(a-b)
= $\dfrac{\cot a \cot b -1}{\cot b +\cot a} \times \dfrac{\cot a \cot b +1}{\cot b -\cot a}$
= $\dfrac{(\cot a \cot b -1)(\cot a \cot b +1)}{(\cot b +\cot a) (\cot b -\cot a)}$
= $\dfrac{(\cot a \cot b)^2 -1^2}{\cot^2 b -\cot^2 a}$, by the algebraic formula: (x-y)(x+y) =x2-y2.
= $\dfrac{\cot^2 a \cot^2 b -1}{\cot^2 b -\cot^2 a}$
So cot(a+b) cot(a-b) formula in terms of cot is given by
cot(a+b) cot(a-b) = $\dfrac{\cot^2 a \cot^2 b -1}{\cot^2 b -\cot^2 a}$.
Also Read: Formula of sin(a+b) sin(a-b)
Cot(a+b) Cot(a-b) Formula in Terms of Tan
As cotx = $\frac{1}{\tan x}$, from the above formula we obtain that
cot(a+b) cot(a-b) = $\dfrac{\dfrac{1}{\tan^2 a \tan^2 b} -1}{\dfrac{1}{\tan^2 b} -\dfrac{1}{\tan^2 a}}$
Simplifying it we get that
cot(a+b) cot(a-b) = $\dfrac{1-\tan^2 a \tan^2 b}{\tan^2 a -\tan^2 b}$
So the formula of cot(a+b)cot(a-b) in terms of tan is given by cot(a+b)cot(a-b) = $\dfrac{1-\tan^2 a \tan^2 b}{\tan^2 a -\tan^2 b}$.
You can read: cosx cosy Formula, Proof
FAQs
Q1: What is the formula of cot(a+b)cot(a-b)?
Answer: The formula of cot(a+b)cot(a-b) is equal to cot(a+b) cot(a-b) = (cot2a cot2b -1)/(cot2b -cot2a).