CosA/1+sinA=1-sinA/cosA
I have to prove the identity each side has to b the same how would i do that? I have a unit test on monday so im doing practice questions
Just cross-multiply to clear the fractions. You get
cos*cos = (1-sin)(1+sin)
If that doesn't look true, you do need to work some more of these. Dozens of them.
sin^2 + cos^2 = 1
is one of the most useful identities you have for trig problems.
Yes but can u help me do one full one so i can understand how to do it. And u cant cros multiply because there is a equal sign between them and our teacher said just to work with the complicated side and then work it out until u get the right side so how would i do that with this one and the one i asked before please help me out!!! Thanksss a bunch sir😃
OK, but the effort is the same.
cos/(1+sin)
Now, you know that a^2-b^2 = (1+a)(1-a), and you knwo that cos^2 = 1-sin^2, so that should lead you to simplify the denominator by multiplying top and bottom by (1-sin)
cos/(1+sin) * (1-sin)/(1-sin)
= cos(1-sin)/(1-sin^2)
= cos(1-sin)/cos^2
= (1-sin)/cos
Time to dredge up your algebra I skills to handle the algebra part, after plugging in your trig identities
In this problem, both sides are equally complicated. Many times there is no clear way to get from one side to the other, and it's perfectly clear to manipulate and simplify both sides independently, with the goal of arriving at the same expression on both sides. At that point you're done, since you haven't changed the actual value of either side, and they wind up being the same expression.
Ohhhhhhhhhhhhhhhhhhhhhh omg!!! Thankyouuuuu sooooo much that helped me a lot!! Ur the best👍so cool:D thankyouuu!
To prove an identity, such as the given equation (CosA/1+sinA = 1-sinA/cosA), you need to manipulate one side of the equation to get it to match the other side.
Start by working with the left-hand side (LHS) of the equation:
CosA / (1 + sinA)
One approach is to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which in this case is (1 - sinA). This can be written as:
((CosA / (1 + sinA)) * ((1 - sinA) / (1 - sinA)))
Simplifying this expression, we get:
((CosA * (1 - sinA)) / ((1 + sinA) * (1 - sinA)))
Further simplifying, we have:
(CosA - CosA * sinA) / (1 - sin^2A)
Using the Pythagorean identity sin^2A = 1 - cos^2A, we can rewrite this as:
(CosA - CosA * sinA) / cos^2A
Factoring outcosA, we get:
CosA * (1 - sinA) / cos^2A
Canceling out the common factor of cosA, we have:
(1 - sinA) / cosA
Now, let's manipulate the right-hand side (RHS) of the equation:
(1 - sinA) / cosA
We can see that the LHS and RHS now match, so we have proved the identity:
CosA / (1 + sinA) = (1 - sinA) / cosA
Remember to check that the manipulations and substitutions used to derive the identity are only valid within the domain and range of the trigonometric functions involved.