prove that
sinA-cosA+1/sinA+cosA-1=1/secA-tanA
It is not clear whether the denominator is sin A or (sin A + cos A) or (sin A + cos A -1) on the left side. It is also not clear whether the denominator is sec A or (sec A - tan A) on the right side. Please use parentheses to clarify the meaning of typed equations. Otherwise we can't help you.
To prove the given equation sinA - cosA + 1/sinA + cosA - 1 = 1/secA - tanA, we need to simplify both sides of the equation until they are equal.
Let's start by simplifying the left side of the equation:
sinA - cosA + 1/sinA + cosA - 1
Combine the like terms:
(sinA + cosA) + 1/sinA - 1
To add the fractions, we need to find a common denominator. The common denominator here is sinA. Multiplying the second fraction by sinA/sinA:
[(sinA + cosA) + sinA/sinA - 1]
Simplifying the fractions:
[(sinA + cosA) + (sinA^2/sinA) - 1]
Using the identity sin^2(A) + cos^2(A) = 1, we can replace sin^2(A) with (1 - cos^2(A)):
[(sinA + cosA) + ((1 - cos^2(A))/sinA) - 1]
Further simplification:
[(sinA + cosA) + (1 - cos^2(A))/sinA - 1]
[(sinA + cosA) - (cos^2(A) - 1)/sinA]
Factoring the numerator of the second fraction as a difference of squares:
[(sinA + cosA) - ((cosA + 1)(cosA - 1))/sinA]
Now, let's simplify the right side of the equation:
1/secA - tanA
SecA is the reciprocal of cosA, so 1/secA is cosA:
cosA - tanA
Writing tanA as sinA/cosA:
cosA - sinA/cosA
To add these fractions, we need a common denominator, which is cosA:
(cosA^2 - sinA)/cosA
Using the identity sin^2(A) + cos^2(A) = 1, we can replace sin^2(A) with (1 - cos^2(A)):
(cos^2(A) - (1 - cos^2(A)))/cosA
Further simplification:
(2cos^2(A) - 1)/cosA
We can rewrite 2cos^2(A) - 1 as cos(2A) using the double-angle identity.
(cos(2A))/cosA
Using the identity cos(A)/cos(B) = sec(B)/sec(A):
sec(2A)/sec(A)
Now let's compare the left side and the right side of the equation we have obtained:
[(sinA + cosA) - ((cosA + 1)(cosA - 1))/sinA] = sec(2A)/sec(A)
To proceed further, we need to use trigonometric identities to simplify both sides and show that they are equal.