if tan a +sin a= x
tan a -sin a= y
show x square - y square=4 rootxy
x^2 = tan^2 +2 tan sin + sin^2
y^2 = tan^2 -2 tan sin + sin^2
x^2-y^2 = 4 tan sin = 4 sin^2/cos
____________________________________
x y = tan^2 -sin^2 = sin^2/cos^2 -sin^2
= sin^2/cos^2 - sin^2 cos^2/cos^2
= [sin^2 / cos^2] [1-cos^2]
but 1 - cos^2 = sin^2
so
= sin^4/cos^2
so
sqrt(xy) = sin^2/cos
and sure enough
4 sqrt (xy) = 4 sin^2/cos
or, try this to start out
x+y = 2tan
x-y = 2sin
x^2-y^2 = (x+y)(x-y) = 4tan sin = 4sin^2/cos
Saves having to expand those pesky squared binomials.
but much more fun my way :)
To solve for x^2 - y^2 and show that it equals 4√xy, we need to manipulate the given equations and use trigonometric identities.
Given:
tan a + sin a = x ----(Equation 1)
tan a - sin a = y ----(Equation 2)
Step 1: Squaring both sides of Equation 1
(tan a + sin a)^2 = x^2
tan^2 a + 2tan a sin a + sin^2 a = x^2
Step 2: Squaring both sides of Equation 2
(tan a - sin a)^2 = y^2
tan^2 a - 2tan a sin a + sin^2 a = y^2
Step 3: Adding Equation 1 and Equation 2 together
(tan^2 a + 2tan a sin a + sin^2 a) + (tan^2 a - 2tan a sin a + sin^2 a) = x^2 + y^2
2tan^2 a + 2sin^2 a = x^2 + y^2 ----(Equation 3, as tan^2 a cancels out and 2sin^2 a = 2 - 2cos^2 a)
Step 4: Simplifying Equation 3 using trigonometric identity
2(tan^2 a + sin^2 a) = x^2 + y^2
2 = x^2 + y^2
Step 5: Solving for x^2 - y^2
x^2 - y^2 = (x^2 + y^2) - 2y^2
= 2 - 2y^2
= 2(1 - y^2)
Step 6: Using trigonometric identity tan^2 a - sin^2 a = 1
x^2 - y^2 = 2(1 - y^2)
= 2(tan^2 a - sin^2 a - sin^2 a) [Replacing 1-y^2 with tan^2 a - sin^2 a using tan^2 a - sin^2 a = 1]
= 2tan^2 a - 4sin^2 a
= 2tan^2 a - 4(1 - cos^2 a) [Replacing sin^2 a with 1 - cos^2 a using sin^2 a + cos^2 a = 1]
= 2tan^2 a - 4 + 4cos^2 a
= 2(tan^2 a + 2cos^2 a) - 4
Step 7: Using trigonometric identity tan^2 a + 2cos^2 a = 2
x^2 - y^2 = 2(2) - 4
= 4 - 4
= 0
Hence, x^2 - y^2 = 0, not 4√xy. Please double-check your equations or confirm if you made any mistakes while writing the problem.