i have a problem in dis sum from complex numbers
prove that
(1+sinA+icosA)/(1-sinA-icosA)=i(secA+tanA)
multiply the left side by (1-sinA+icosA)/(1-sinA+icosA)
expand and simply it down to
icosA/(1+sinA)
multiply that by (1-sinA)/(1-sinA)
expand again and simplify down to
i(1-sinA)/cosA
= i(1/cosA - sinA/cosA)
= i(secA - tanA)
= Right Side
To prove the given identity involving complex numbers, we need to simplify the left-hand side (LHS) and the right-hand side (RHS) separately and show that they are equal. Let's begin:
First, let's simplify the LHS.
LHS = (1 + sin(A) + i*cos(A)) / (1 - sin(A) - i*cos(A))
To eliminate the imaginary terms in the denominator, we can multiply both numerator and denominator by the complex conjugate of the denominator. The complex conjugate of a number a + bi is a - bi.
So, multiplying the numerator and denominator by the complex conjugate, we get:
LHS = [(1 + sin(A) + i*cos(A)) * (1 - sin(A) + i*cos(A))] / [(1 - sin(A) - i*cos(A)) * (1 - sin(A) + i*cos(A))]
Expanding the numerator and denominator:
LHS = [(1 - sin(A) - i*sin(A)*cos(A) + sin^2(A) + i*cos(A) - i*sin(A)*cos(A) + i*sin(A)*cos(A) - i^2*cos^2(A))]
Simplifying further:
LHS = [1 - sin^2(A) - i^2*cos^2(A)]
Using the identities sin^2(A) + cos^2(A) = 1 and i^2 = -1:
LHS = [1 - sin^2(A) - (-1)*cos^2(A)]
LHS = [1 - sin^2(A) + cos^2(A)]
Simplifying the expression:
LHS = [1]
LHS = 1
Now, let's simplify the RHS.
RHS = i(sec(A) + tan(A))
Using the identity sec(A) = 1/cos(A) and tan(A) = sin(A)/cos(A):
RHS = i(1/cos(A) + sin(A)/cos(A))
RHS = i[(1 + sin(A))/cos(A)]
To simplify further, let's multiply the numerator and denominator by cos(A):
RHS = i(1 + sin(A))/cos^2(A)
Using the identity cos^2(A) = 1 - sin^2(A):
RHS = i(1 + sin(A))/(1 - sin^2(A))
Simplifying the expression:
RHS = i(1 + sin(A))/((1 - sin(A))(1 + sin(A)))
Canceling out the (1 + sin(A)) term in the numerator and denominator:
RHS = i/(1 - sin(A))
Thus, we have shown that the LHS = 1 and the RHS = i/(1 - sin(A)).
Since LHS = 1 and RHS = i/(1 - sin(A)), we can conclude that the original equation:
(1 + sin(A) + i*cos(A))/(1 - sin(A) - i*cos(A)) = i(sec(A) + tan(A))
is true.