Prove that (cos2A)÷ ( 1+sin2A) = (cotA -1) ÷ (cotA +1)
The trick here is to recall that 1 = cos^a+sin^2A
cos2A/(1+sin2A)
(cos^2A-sin^2A)/(1+2sinAcosA)
(cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A)
(cosA+sinA)(cosA-sinA)/(cosA+sinA)^2
(cosA-sinA)/(cosA+sinA)
(cotA-1)/(cotA+1)
The trick here is to recall that 1 = cos^a+sin^2A
cos2A/(1+sin2A)
(cos^2A-sin^2A)/(1+2sinAcosA)
(cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A)
(cosA+sinA)(cosA-sinA)/(cosA+sinA)^2
(cosA-sinA)/(cosA+sinA)
(cotA-1)/(cotA+1)
To prove that (cos2A) ÷ (1 + sin2A) = (cotA - 1) ÷ (cotA + 1), we need to manipulate both sides of the equation using trigonometric identities to obtain the same result.
Let's start by recalling some common trigonometric identities:
1. cos2A = cos^2(A) - sin^2(A)
2. sin2A = 2sin(A)cos(A)
3. cotA = cosA/sinA
Now, let's proceed with the proof:
Left-hand side (LHS):
(cos2A) ÷ (1 + sin2A)
Using the identity cos2A = cos^2(A) - sin^2(A):
= (cos^2(A) - sin^2(A)) ÷ (1 + sin2A)
Using the identity sin2A = 2sin(A)cos(A):
= (cos^2(A) - sin^2(A)) ÷ (1 + 2sin(A)cos(A))
Using the identity cos^2(A) = 1 - sin^2(A):
= ((1 - sin^2(A)) - sin^2(A)) ÷ (1 + 2sin(A)cos(A))
= (1 - 2sin^2(A)) ÷ (1 + 2sin(A)cos(A))
Now, let's simplify the right-hand side (RHS):
(cotA - 1) ÷ (cotA + 1)
Using the identity cotA = cosA/sinA:
= (cosA/sinA - 1) ÷ (cosA/sinA + 1)
To simplify this, we need to manipulate the numerator by multiplying it by (sinA - cosA) and the denominator by (sinA + cosA):
= [(cosA - sinA)/(sinA)] ÷ [(cosA + sinA)/(sinA)]
Simplifying further:
= (cosA - sinA)/(sinA) * (sinA)/(cosA + sinA)
= (cosA - sinA)/(cosA + sinA)
Now, we can see that the LHS = RHS:
(1 - 2sin^2(A)) ÷ (1 + 2sin(A)cos(A)) = (cosA - sinA)/(cosA + sinA)
Hence, we have proven that (cos2A) ÷ (1 + sin2A) = (cotA - 1) ÷ (cotA + 1) using trigonometric identities.