A community for students.
Here's the question you clicked on:
 0 viewing
vera_ewing
 one year ago
@Michele_Laino
vera_ewing
 one year ago
@Michele_Laino

This Question is Closed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1applying the distrinutive property of multiplication over addition, we can write the se steps: \[\Large \begin{gathered} {\left( {\cos \theta } \right)^2}\left\{ {1 + {{\left( {\tan \theta } \right)}^2}} \right\} = \hfill \\ \hfill \\ = {\left( {\cos \theta } \right)^2} + {\left( {\cos \theta } \right)^2}{\left( {\tan \theta } \right)^2} = \hfill \\ \hfill \\ = {\left( {\cos \theta } \right)^2} + {\left( {\cos \theta } \right)^2}\frac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: dw:1436196847953:dw

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Oh so the answer must be A!

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait I got it wrong :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please keep in mind that: \[\Large \begin{gathered} {\left( {\cos \theta } \right)^2} + {\left( {\cos \theta } \right)^2}\frac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \hfill \\ \hfill \\ = {\left( {\cos \theta } \right)^2} + {\left( {\sin \theta } \right)^2} = 1 \hfill \\ \end{gathered} \]

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Yep. That makes sense. Thanks Michele.
Ask your own question
Sign UpFind 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.