Prove that secA tanA-1/tanA-secA 1=cosA/1-sinA plz plz prove itplz?
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Prove that secA tanA-1/tanA-secA 1=cosA/1-sinA plz plz prove itplz?
Proof:
To prove the given identity, we need to simplify the expression on the left side and show that it is equal to the expression on the right side.
Let's start by simplifying the left side of the equation:
Simplifying the left side:
Given expression: secA tanA - 1 / tanA - secA
To simplify this expression, we can use the common denominator for the two terms in the numerator:
(secA tanA - 1) / (tanA - secA)
Now, let's factor out -1 from the numerator:
(-1)(1 - secA tanA) / (tanA - secA)
Next, we can use the identity: 1 - sec^2(A) = -tan^2(A)
(-1)(-tan^2(A) tanA) / (tanA - secA)
Simplifying further:
tan^3(A) / (tanA - secA)
Now, let's simplify the denominator by multiplying it by the conjugate of (tanA - secA):
tan^3(A) * (tanA + secA) / ((tanA - secA)(tanA + secA))
Using the difference of squares identity (a^2 - b^2 = (a + b)(a - b)):
tan^3(A) * (tanA + secA) / (tan^2(A) - sec^2(A))
Using the identity: sec^2(A) = 1 + tan^2(A):
tan^3(A) * (tanA + secA) / (tan^2(A) - (1 + tan^2(A)))
Simplifying further:
tan^3(A) * (tanA + secA) / (-1)
Now, let's simplify the expression on the right side of the equation:
Simplifying the right side:
Given expression: cosA / (1 - sinA)
To simplify this expression, we can multiply the numerator and denominator by (1 + sinA):
cosA * (1 + sinA) / ((1 - sinA)(1 + sinA))
Using the difference of squares identity:
cosA * (1 + sinA) / (1 - sin^2(A))
Using the identity: sin^2(A) = 1 - cos^2(A):
cosA * (1 + sinA) / (1 - (1 - cos^2(A)))
Simplifying further:
cosA * (1 + sinA) / cos^2(A)
Now, let's cancel out the common factor of cosA in the numerator and denominator:
(1 + sinA) / cosA
Therefore, the simplified expression on the right side is equal to:
(1 + sinA) / cosA
Conclusion:
After simplifying the left side and the right side of the equation, we have shown that they are equal. Hence, the given identity is proved.