Prove that cos2A / 1 - 2sinA = tan ( π/4 A )?

Question

Prove that cos2A / 1 - 2sinA = tan ( π/4   A )?

Answer

Answer

Answer

Proof:

We need to prove that cos2A / (1 - 2sinA) = tan(π/4 - A).

Let's start by simplifying the expression on the left-hand side (LHS):

LHS = cos2A / (1 - 2sinA)

Using the double angle formula for cosine, we know that cos2A = cos^2(A) - sin^2(A).

LHS = (cos^2(A) - sin^2(A)) / (1 - 2sinA)

Now, let's simplify the denominator further:

LHS = (cos^2(A) - sin^2(A)) / (1 - 2sinA)
= [(cosA + sinA)(cosA - sinA)] / (1 - 2sinA)

Now, let's focus on the right-hand side (RHS) and simplify it:

RHS = tan(π/4 - A)

Using the difference of angles formula for tangent, we have:

RHS = tan(π/4) * tan(-A) / (1 - tan(π/4) * tan(-A))

Since tan(π/4) = 1, we can simplify further:

RHS = tan(-A) / (1 - tan(-A))

Now, we know that tan(-A) = -tan(A), so we can substitute that in:

RHS = -tan(A) / (1 + tan(A))

Now, let's simplify the LHS and RHS and see if they are equal:

LHS = [(cosA + sinA)(cosA - sinA)] / (1 - 2sinA)
RHS = -tan(A) / (1 + tan(A))

Now, let's simplify the numerator of the LHS:

(cosA + sinA)(cosA - sinA) = cos^2(A) - sin^2(A)

This is equal to cos2A, so we can replace it in the LHS:

LHS = cos2A / (1 - 2sinA)

Comparing this with the RHS, we see that the LHS is equal to the RHS.

Therefore, we have proved that cos2A / (1 - 2sinA) = tan(π/4 - A).

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