I am trying to help my daughter with her trig and this one has me stumped.
Verify the following identity. Work on only one side of the equation to get it to equal the other. You may NOT cross mutlply.
csc x +sec x/sin x + cos x = cot x + tan x
Just to verify, if you were going to place parenthesis on the left hand side of the equation, where would you place them?
The equation as stated is not correct, so it’s impossible to solve. Is there something you’re leaving out?
Are you sure you copied the problem correctly?
If not, try plugging in some value in to check if it actually holds (45º is good). Because it doesn’t work as written (parentheses or not), and there could be a problem with the assignment.
I think you can add parentheses to make it valid:
(csc x +sec x)/(sin x + cos x) = cot x + tan x
Props to MissMossie for pointing the way.
Yes, punoqllads has the parenthesis correct. But I still can’t solve it.
Do you want the answer or just a hint?
Hints:
Start with the most complicated side and make it simpler(this applies to solving all such identities)
Write everything in terms of cos and sin only(again, this applies to any identity)
Get rid of multiple “levels” of fractions (eg 1/((cot x) / (sin x + cos x)) + tan x has multiple levels). Reduce it all to a single fraction with one numerator and one denominator
Even though the problem says you should only change one side, you can usually work on both sides to reduce them to the same thing, and than write up the proof so you reduce one side to that thing and then reverse the reduction of the other side on it to get the other side back – if that made any sense at all.
If you’ve made one side as simple as possible and you don’t know where to go from there, try making it more complicated instead.
The answer:
(csc x + sec x)/(sin x + cos x)
= (1/sin x + 1/cos x)/(sin x + cos x)
= (sin x + cos x)/(sin x * cos x)/(sin x + cos x)
= (sin x + cos x)/((sin x * cos x)*(sin x + cos x))
= 1 / (sin x * cos x)
= (sin^2 x + cos^2 x)/(sin x * cos x)
= sin^2 x/(sin x * cos x) + cos^2 x/(sin x * cos x)
= sin x / cos x + cos x / sin x
= tan x + cot x
At least my thoughts that the problem was wrong as stated was verified. It’s a very different problem now.
One little bit of help that I hope will point you on the right path: csc(x) is equal to 1/sin(x), and sec(x) is equal to 1/cos(x). So since sec(x) + csc(x) = 1/sin(x) + 1/cos(x), you can put them together with a common denominator of sin(x) cos(x): sec(x) + csc(x) = cos(x)/sin(x)cos(x) + sin(x)/sin(x)cos(x) = (cos(x) + sin(x))/sin(x)cos(x) = (sin(x) + cos(x))/sin(x)cos(x).
Does that help any?
there are certain equivalents when dealing with sin cos csc sec it will get UGLY just get a TI-89, it will practically solve it for you, plus your daughter will need it for calc
If his daughter’s in high school, she’ll be expected to solve trig identities without a calculator. They’re all solvable by hand, especially if you use the general guidelines Rysto outlined above, keep the reciprocal and pythagorean identities in mind, and try standard Algebra I techniques like factoring and cross-multiplying.
Rysto gave some very good advice, and a correct solution.
For what it’s worth, the solution I found doesn’t (directly) involve writing everything in terms of sin and cos:
First, multiply numerator and denominator of the left side by (csc x – sec x). With a little algebra, it turns into
(csc[sup]2[/sup] x – sec[sup]2[/sup] x)/(cot x – tan x)
Now use the identities sec[sup]2[/sup] x = tan[sup]2[/sup] x + 1 and csc[sup]2[/sup] x = cot[sup]2[/sup] + 1 on the numerator.
Finally, factor the numerator (difference of two squares) and cancel the common factor.
Every trig identity involving a single angle can be resolved by first replacing tan, cot, sec, csc by their definitions in terms of sin and cos and by using the fact that sin^2 + cos^2 = 1. Applied in the current case, both sides reduce to the same place: 1/sin*cos and you can obviously turn things around to get left hand side = right hand side.
Incidentally, there is an amusing identity hidden here:
(1/a + 1/b)/(a + b) = 1/ab.