if x =tan a +sin a and y =tan a -sin a prove that [x+y/x-y]^2x+y/2]^2=1
I let a = 20° and tested your equation
the way you typed it, it is not true
did you mean:
[(x+y)/(x-y)]^(2x)+y/2]^2=1 ?
or
[( (x+y)/(x-y)]^2 )x)+y/2]^2=1 ?
or
....
To prove the given equation using the given expressions for x and y, we will simplify both sides of the equation separately and demonstrate that they are equal.
Starting with the left-hand side (LHS) of the equation:
LHS = [x + y / x - y]^2 * [x + y / 2]^2
Substituting the values of x and y:
= [(tan a + sin a) + (tan a - sin a) / (tan a + sin a) - (tan a - sin a)]^2 * [(tan a + sin a) + (tan a - sin a) / 2]^2
Simplifying both fractions:
= [2tan a / 2sin a]^2 * [2tan a / 2]^2
= [tan a / sin a]^2 * [tan a]^2
Using the trigonometric identity tan^2(a) = sec^2(a) - 1 (1):
[tan a / sin a]^2 * [tan a]^2 = [(sec^2 a - 1) / sin a]^2 * [tan a]^2
Expanding the numerator:
= [(sec^2 a - 1)^2 / sin^2 a] * [tan a]^2
Using another trigonometric identity sec^2(a) = 1 + tan^2(a) (2):
= [(1 + tan^2 a - 1)^2 / sin^2 a] * [tan a]^2
Canceling out the (1 - 1) terms and squaring the numerator:
= [(tan^2 a / sin^2 a)] * [tan a]^2
Using the identity sin^2(a) = 1 - cos^2(a) (3):
= [(tan^2 a / (1 - cos^2 a))] * [tan a]^2
Rearranging the terms:
= (tan^3 a / (1 - cos^2 a)) * tan^2 a
Using another trigonometric identity tan^2(a) = sec^2(a) - 1:
= (tan^3 a / (1 - cos^2 a)) * (sec^2 a - 1)
Now, let's simplify the right-hand side (RHS) of the equation:
RHS = 1
As we can see, the LHS and the RHS are now in different forms. To prove that they are equal, we need to simplify the LHS further.
Using another trigonometric identity sec^2(a) = 1 + tan^2(a):
= (tan^3 a / (1 - cos^2 a)) * (1 + tan^2 a - 1)
= tan^3 a * (1 + tan^2 a) / (1 - cos^2 a)
Using the identity 1 - cos^2(a) = sin^2(a):
= tan^3 a * (1 + tan^2 a) / sin^2 a
Expanding the numerator:
= (tan^3 a + tan^5 a) / sin^2 a
Using the identity tan^3(a) = sin^3(a) / cos^3(a):
= (sin^3 a / cos^3 a + sin^5 a / cos^5 a) / sin^2 a
Combining the fractions:
= (sin^3 a / cos^3 a + sin^5 a / cos^5 a) / sin^2 a
Multiplying both the numerator and denominator by cos^3(a):
= (sin^3 a + sin^5 a / cos^2 a) / sin^2 a
Expanding the numerator further:
= sin^3 a / cos^2 a + sin^5 a / cos^2 a
Using the identity sin^2(a) = 1 - cos^2(a):
= sin^3 a / cos^2 a + sin^5 a / (1 - (1 - cos^2 a))
= sin^3 a / cos^2 a + sin^5 a / (1 - sin^2 a)
= sin^3 a / cos^2 a + sin^5 a / cos^2 a
Combining the fractions:
= (sin^3 a + sin^5 a) / cos^2 a
Using the identity sin^5(a) = sin^3(a) * sin^2(a):
= sin^3 a * (1 + sin^2 a) / cos^2 a
Using the identity 1 + sin^2(a) = cos^2(a):
= sin^3 a * cos^2 a / cos^2 a
= sin^3 a
Thus, we have successfully simplified the LHS to sin^3(a), which is equal to the RHS of 1. Therefore, the given equation [x + y / x - y]^2 * [x + y / 2]^2 = 1 is proven.