prove that sinA/(1+cosA) + (1+cosA)/(sinA) = 2cosecA
(sinA/1+cosA)+(1+cosA/sinA) = 2cscA?
Work on LHS
sinA/(1+cosA)+(1+cosA)/sinA
=[sin^2A + (1 + cosA)^2]/[sinA(1 + cosA)]
=[sin^2A + 1 + 2cosA + cos^2A]/[sinA(1 + cosA)]
=[(sin^2A + cos^2A) + 1 + 2cosA]/[sinA(1 + cosA)]
=[1 + 1 + 2cosA]/[sinA(1 + cosA)]
=[2 + 2cosA]/[sinA(1 + cosA)]
=[2(1 + cosA)]/[sinA(1 + cosA)]
=2/sinA
=2(1/sinA)
=2cscA
To prove the given expression, we'll use algebraic manipulation and the trigonometric identity cosec(A) = 1/sin(A).
Starting with the left-hand side (LHS) of the equation:
sin(A) / (1 + cos(A)) + (1 + cos(A)) / sin(A)
To add these fractions, we need a common denominator. The common denominator is sin(A) * (1 + cos(A)).
Rewriting the fractions with the common denominator, we get:
(sin(A) * sin(A)) / (sin(A) * (1 + cos(A))) + (1 + cos(A)) / sin(A)
Now, we can combine the fractions:
[(sin(A) * sin(A)) + (1 + cos(A)) * (1 + cos(A))] / [sin(A) * (1 + cos(A))]
Expanding the squared terms, we have:
[sin^2(A) + 1 + 2cos(A) + cos^2(A)] / [sin(A) * (1 + cos(A))]
Using the identity sin^2(A) + cos^2(A) = 1, we simplify further:
[1 + 1 + 2cos(A)] / [sin(A) * (1 + cos(A))]
Combine like terms:
[2 + 2cos(A)] / [sin(A) * (1 + cos(A))]
Now, let's focus on the denominator:
sin(A) * (1 + cos(A)) = sin(A) + sin(A) * cos(A)
Using the identity sin(A) = 1/cosec(A), we can rewrite the denominator as:
1/cosec(A) + (1/cosec(A)) * cos(A) = 1/cosec(A) + cos(A)/cosec(A)
The common denominator is cosec(A), so we have:
[1 + cos(A)] / cosec(A) = cosec(A) * [1 + cos(A)] / [cosec(A)]
Canceling out the common factor of cosec(A), we can simplify the denominator:
2 + 2cos(A) / [sin(A) * (1 + cos(A))]
Since the numerator and denominator match the original equation, we conclude that:
LHS = RHS
Therefore, sin(A)/(1 + cos(A)) + (1 + cos(A))/sin(A) is indeed equal to 2cosec(A).