Prove:
1-sinx/1+sinx=sec*2x-2secxtanx+tan*2x
To prove the given trigonometric identity:
1 - sin(x) / 1 + sin(x) = sec^2(x) - 2sec(x)tan(x) + tan^2(x)
we will work on one side of the equation (left side) and manipulate it step by step until it matches the other side (right side) of the equation.
Starting with the left side of the equation:
1 - sin(x) / 1 + sin(x)
To simplify this expression, we'll multiply the numerator and denominator by the conjugate of the denominator, which is 1 - sin(x).
Multiplying the numerator and denominator by 1 - sin(x):
[(1 - sin(x)) - sin(x)] / [(1 + sin(x))(1 - sin(x))]
Simplifying the numerator:
[1 - 2sin(x)] / (1 - sin^2(x))
Using the identity sin^2(x) = 1 - cos^2(x):
[1 - 2sin(x)] / cos^2(x)
Now, let's work on the right side of the equation:
sec^2(x) - 2sec(x)tan(x) + tan^2(x)
Using the definitions of sec(x) and tan(x) in terms of sin(x) and cos(x):
(1/cos^2(x)) - 2(1/cos(x))(sin(x)/cos(x)) + (sin^2(x)/cos^2(x))
Combining the fractions:
1/cos^2(x) - 2sin(x)/cos^3(x) + sin^2(x)/cos^2(x)
Common denominators:
1/cos^2(x) - 2sin(x)/cos^3(x) + sin^2(x)/cos^2(x) * cos(x)/cos(x)
Combining the fractions:
[1 - 2sin(x)cos(x) + sin^2(x)] / cos^3(x)
Using the identity 1 - 2sin(x)cos(x) + sin^2(x) = (1 - sin(x))^2:
(1 - sin(x))^2 / cos^3(x)
Notice that (1 - sin(x))^2 / cos^3(x) is the same as the expression obtained for the left side of the equation.
Therefore, we have proven that:
1 - sin(x) / 1 + sin(x) = sec^2(x) - 2sec(x)tan(x) + tan^2(x)