It has been proven that the trigonometric Identity (1 - secA + tanA)/(1 + secA + TanA) is equal to; (secA + tanA - 1)/(secA + tanA + 1)
We want to prove that;
(1 - secA + tanA)/(1 + secA + TanA) = (secA + tanA - 1)/(secA + tanA + 1)
We know from trigonometric identities that;
(secA)² – (tanA)² = 1 ----(1)
Also, from algebra we know that;
a² - b² = (a + b)(a - b)
The numerator of LHS of the original given Identity is:
1 - sec A - tan A
Using equation 1, we can say that;
(secA)² – (tanA)² - (sec A + tan A)
⇒ (sec A + tan A)(sec A - tan A) - (sec A - tan A)
This can be factorized to get;
(sec A - tan A)(sec A + tan A - 1)
Similarly, the denominator can be expressed as;
(sec A - tan A)(sec A + tan A + 1)
Thus, combining the numerator and denominator together gives us:
[(sec A - tan A)(sec A + tan A - 1)]/[(sec A - tan A)(sec A + tan A + 1)]
⇒ (secA + tanA - 1)/(secA + tanA + 1)
That expression is equal to the Right hand side and as such the Trigonometric Identity is proved.
Read more about Trigonometric Identities at; https://brainly.com/question/22591162
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