4m=cotA(1+SinA), 4n=cotA(1-SinA)
Proved that
(m2-n2)=mn
There must be a typo. Plug in A = pi/4 and you have
4m = (1 + 1/√2), so m = (2+√2)/8
4n = (1 - 1/√2), so n = (2-√2)/8
m2-n^2 = √2/8
mn = 1/32
To prove that (m^2 - n^2) = mn using the given equations, we need to manipulate the equations and apply trigonometric identities to simplify and relate the expressions.
Given:
4m = cotA(1 + sinA) -- Equation 1
4n = cotA(1 - sinA) -- Equation 2
Let's start by manipulating Equation 1:
4m = cotA(1 + sinA)
Rearrange by dividing both sides by cotA:
4m/cotA = 1 + sinA
Simplify the left side by using the trigonometric identity cotA = 1/tanA:
4m/(1/tanA) = 1 + sinA
Multiply both sides by tanA to eliminate the fraction:
4m * tanA = tanA + sinA * tanA
Using the identity tanA = sinA / cosA, we can simplify further:
4m * (sinA / cosA) = (sinA / cosA) + (sinA * sinA / cosA)
Combine terms on the right side:
4m * sinA / cosA = (sinA + sin^2 A) / cosA
Rearrange and expand the right side using the identity sin^2 A + cos^2 A = 1:
4m * sinA / cosA = (1 - cos^2 A + sin^2 A) / cosA
Simplify:
4m * sinA = (1 - cos^2 A + sin^2 A)
Recognize that 1 - cos^2 A = sin^2 A (using the identity sin^2 A + cos^2 A = 1):
4m * sinA = sin^2 A + sin^2 A
Combine like terms:
4m * sinA = 2sin^2 A
Now, let's manipulate Equation 2:
4n = cotA(1 - sinA)
Using similar steps as before, we'll arrive at:
4n * sinA = 2sin^2 A
Now, we can compare the two equations:
4m * sinA = 2sin^2 A
4n * sinA = 2sin^2 A
Divide both sides of each equation by 2sin^2 A:
(4m * sinA) / (2sin^2 A) = 1
(4n * sinA) / (2sin^2 A) = 1
Simplify:
2m / sinA = 1
2n / sinA = 1
Rearrange each equation:
2m = sinA
2n = sinA
Now, we can substitute sinA in the equation (m^2 - n^2) = mn:
(m^2 - n^2) = mn
(m^2 - n^2) = (2m * 2n)
Simplify:
(m^2 - n^2) = (4mn)
Now, let's replace sinA in terms of m and n:
(4mn) = (4mn)
Therefore, we have proved that (m^2 - n^2) = mn using the given equations.