cot²A-cot²B = (cos²A - cos²B)/(sin²A sin²B)
Answers
Step-by-step explanation:
I think you have to prove this, let's start:
We know CotA=cosA/Sina
=>Cot^2A=Cos^2A/Sin^2A
Let's use this in the main equation:
=>Cot^2A-Cot^2B = Cos^2A/Sin^2A - Cos^2B/Sin^2B
We need Sin^2A.Sin^2B in the denominator so let's take LCM of both terms.
=>Cos^2A/Sin^2A - Cos^2B/Sin^2B
= (Cos^2A.Sin^2B - Cos^2B.Sin^2A)/(Sin^2A.Sin^2B)
Now we have the required denominator, so let's modify the numerator.
We know, Sin^2A= 1-Cos^2A
Using this for both Sin^2A and Sin^2B in the numerator.
Let's take one term at a time
=>Cos^2A.Sin^2B = Cos^2A.(1-Cos^2B) [multiplying]
=Cos^2A - Cos^2A.Cos^2B .... 1
=>Cos^2B.Sin^2A = Cos^2B.(1-Cos^2A) [multiplying]
=Cos^2B - Cos^2B.Cos^2A .... 2
Apply 1 and 2 in the main equation
(Cos^2A.Sin^2B - Cos^2B.Sin^2A)/(Sin^2A.Sin^2B)
= (Cos^2A-Cos^2A.Cos^2B -[Cos^2B - Cos^2B.Cos^2A])/Sin^2A.Sin^2B
= (Cos^2A-Cos^2B+Cos^2A.Cos^2B-Cos^2A.Cos^2B)/ Sin^2A.Sin^2B
Cancel Cos^2A.Cos^2B-Cos^2A.Cos^2B
=>(Cos^2A-Cos^2B)/Sin^2A.Sin^2B
Hope it helps...
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