A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \frac{1}{1+\sin x}+\frac{1}{\csc x+1}\] Let's do some manipulation. We're going to take advantage of two of the three basic square identities that you first learn in trig.\[\large \color{royalblue}{\sin^2x+\cos^2x=1 \qquad \qquad \cot^2x+1=\csc^2x}\] In our problem, let's multiply the first term by the `conjugate` of the bottom.\[\large \color{orangered}{\frac{1\sin x}{1\sin x}}\frac{1}{1+\sin x}+\frac{1}{\csc x+1}\color{orangered}{\frac{\csc x1}{\csc x1}}\] When we multiply conjugates, we end up with the `Difference of squares` on the bottoms.\[\large \frac{1\sin x}{1\sin^2x}+\frac{\csc x1}{\csc^2x1}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Now we can use our blue identities,\[\large \color{royalblue}{\cos^2x=1\sin^2x \qquad \qquad \cot^2x=\csc^2x1}\] Using these identities gives us,\[\large \frac{1\sin x}{\color{royalblue}{1\sin^2x}}+\frac{\csc x1}{\color{royalblue}{\csc^2x1}} \qquad \rightarrow \qquad \frac{1\sin x}{\color{royalblue}{\cos^2x}}+\frac{\csc x1}{\color{royalblue}{\cot^2x}}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Recalling that, \[\large \color{brown}{\cot^2x=\frac{\cos^2x}{\sin^2x}}\]Gives us,\[\large \frac{1\sin x}{\cos^2x}+\frac{\csc x1}{\color{brown}{\dfrac{\cos^2x}{\sin^2x}}}\]Now notice that we're dividing by a fraction, which we can rewrite as multiplication if we `flip` the bottom fraction.\[\large \frac{1\sin x}{\cos^2x}+\frac{\color{brown}{\sin^2x}(\csc x1)}{\color{brown}{\cos^2x}}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Then distribute the sin^2x to each term in the brackets. There's only a little bit of simplification to do after that. Lemme know if you're still stuck on something.

snowflakepixie
 2 years ago
Best ResponseYou've already chosen the best response.0It took a while to read that since most of the formatting didn't work, but I think I've got it now. Thanks c:

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Formatting didn't work? :c If you refresh the page it usually fixes it. That stinks :d

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.0I feel that the faster way of doing it without any identities is this. \[\csc x=\frac{1}{\sin x}\] \[\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1 }{ \sin x } +1 }\] \[\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1+\sin x }{ \sin x } }\] \[=\frac{ 1 }{ \sin x+1 }+\frac{ \sin x }{1+\sin x}\] \[=\frac{1(1+\sin x+\sin x(\sin x +1)}{(\sin x+1)^2}\] \[=\frac{1+\sin x+\sin^2 x +\sin x}{(\sin x +1)^2}\] \[=\frac{\sin^2 x+2\sin x +1}{(\sin x +1)^2}\] Factorise the numerator \[=\frac{(\sin x +1)^2}{(\sin x +1)^2}\] \[=1\]

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.0\[1(1+\sin x+\sin x(\sin x+1)\] The numerator on one the lines was supposed to have a bracket somewhere. \[1(1+\sin x)+\sin x(\sin x+1)\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.