Question 358824: Verify that...
(Sec(A) - tan(A))^2 = 1-sin(A)/1+sin(A)
Using LHS...I have
sec^2(A)-2sec(A)tan(A) + tan^2(A)
=1-tan^2(A)-2sec(A)tan(A)+tan^2(A)
=1-2sec(A)tan(A)
Now I'm not sure how to proceed.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! (Sec(A) - tan(A))^2 = (1-sin(A))/(1+sin(A))
The problem with your work so far is that sec^2(A) = 1 + tan^2(A), not 1 - tan^2(A).
With these problems, it is often helpful to rewrite sec, csc, tan and cot in terms of sin and/or cos. This is especially true for this problem because your RHS is already in terms of just sin(A). So this is what we will do first:
(1/cos(A) - sin(A)/cos(A))^2 = (1-sin(A))/(1+sin(A))
On the LHS, the two fractions have the same denominator so they can be combined. And since the RHS is just one term, this looks like a good idea:
((1-sin(A))/cos(A))^2 = (1-sin(A))/(1+sin(A))
We can square the fraction on the LHS:
((1-sin(A))^2/cos^2(A) = (1-sin(A))/(1+sin(A))
This looks promising. We already have 1-sin(A) as a factor of the numerator on both sides of the equation. We don't want any cos but, since it's squared, this is easy to turn into an expression of sin(A):
((1-sin(A))^2/(1-sin^2(A)) = (1-sin(A))/(1+sin(A))
What we want is- One of the (1-sin(A)) factors in the numerator on the LHS to "disappear".
- The 1-sin^2(A) in the denominator on the LHS to turn into 1+sin(A) somehow.
As soon as we recognize that 1-sin^2(A) is a difference of squares and that it will factor into (1+sin(A))(1-sin(A)), we will realize that this will take care of both of our goals:
((1-sin(A))^2/((1+sin(A))(1-sin(A)) = (1-sin(A))/(1+sin(A))
On the LHS, (1-sin(A)) is a factor of both the numerator and denominator so they will cancel:
(1-sin(A)/(1+sin(A)) = (1-sin(A))/(1+sin(A))
and we're done.
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