Loading [MathJax]/jax/output/SVG/config.js
 
+0  
 
0
798
2
avatar



Rewrite the expression as a single trigonometric ratio.









 


A)csc θ




 




 


B)1/cos θ




 




 


C)sin θ




 




 


D)sec θ




 






 

 Oct 13, 2014

Best Answer 

 #1
avatar+23254 
+5

sinθ / (1 - cos²θ)

Since   sin²θ + cos²θ  =  1

--->    sin²θ  =  1 - cos²θ

So:  sinθ / (1 - cos²θ)  =  sinθ / sin²θ  =  1 / sinθ  =  cscθ

Trig is easier if you can remember the Pythagorean Identity in its three forms:

     sin²θ + cos²θ  =  1      sin²θ  =  1 - cos²θ          cos²θ  =  1 - sin²θ    

Also, there are two more that derive from the above:

     1 + tan²θ = sec²θ          1 + cot²θ + csc²θ

 Oct 13, 2014
 #1
avatar+23254 
+5
Best Answer

sinθ / (1 - cos²θ)

Since   sin²θ + cos²θ  =  1

--->    sin²θ  =  1 - cos²θ

So:  sinθ / (1 - cos²θ)  =  sinθ / sin²θ  =  1 / sinθ  =  cscθ

Trig is easier if you can remember the Pythagorean Identity in its three forms:

     sin²θ + cos²θ  =  1      sin²θ  =  1 - cos²θ          cos²θ  =  1 - sin²θ    

Also, there are two more that derive from the above:

     1 + tan²θ = sec²θ          1 + cot²θ + csc²θ

geno3141 Oct 13, 2014
 #2
avatar
0

I am confused. I am not sure which answer to pick.

 Oct 13, 2014

2 Online Users

avatar
avatar