Prove that :
[tex] \frac{secA -tanA }{secA +tanA } = 1 - 2secA + 2 {tan}^{2} A[/tex]
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Prove that :
[tex] \frac{secA -tanA }{secA +tanA } = 1 - 2secA + 2 {tan}^{2} A[/tex]
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Sure, let's prove the given trigonometric identity.
We know that secA = 1/cosA and tanA = sinA/cosA.
Let's substitute these values into the left-hand side of the equation:
LHS = (secA - tanA) / (secA + tanA)
= (1/cosA - sinA/cosA) / (1/cosA + sinA/cosA)
= (1 - sinA) / (1 + sinA)
Now, let's use the identity sin^2 A + cos^2 A = 1, which implies that cos^2 A = 1 - sin^2 A.
Substituting this into the LHS, we get:
LHS = (1 - sinA) / (1 + sinA)
= [(1 - sinA)^2 / cos^2 A] / [(1 + sinA)^2 / cos^2 A]
= (1 - 2sinA + sin^4 A) / (1 + 2sinA + sin^4 A)
Now, let's substitute back secA = 1/cosA and tan^2 A = sin^2 A/cos^2 A:
LHS = 1 - 2(1/cosA) + 2(sin^2 A / cos^2 A)
= 1 - 2secA + 2tan^2 A
So, we have proved that:
(secA - tanA) / (secA + tanA) = 1 - 2secA + 2tan^2 A
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