+0  
 
0
871
1
avatar+154 

How do i prove csc2x+sec2x=csc2xsec2x?

I dont even know which identity to start with...

i know csc=1/sin

          sec=1/cos

          sin2+cos2=1

 Jun 6, 2014

Best Answer 

 #1
avatar+33659 
+13

csc = 1/sin and sec = 1/cos so the left-hand side of the equality can be written as:

csc2x + sec2x = 1/sin2x + 1/cos2x

Collect over the common denominator obtained by multiplying sin2x and cos2x together and use the fact that cos2 + sin2 = 1 .

csc2x + sec2x = 1/sin2x + 1/cos2x = (cos2x + sin2x)/(sin2x*cos2x) = 1/(sin2x*cos2x) = csc2x*sec2x

The final expression is the same as the right-hand side of the original equality.

 Jun 6, 2014
 #1
avatar+33659 
+13
Best Answer

csc = 1/sin and sec = 1/cos so the left-hand side of the equality can be written as:

csc2x + sec2x = 1/sin2x + 1/cos2x

Collect over the common denominator obtained by multiplying sin2x and cos2x together and use the fact that cos2 + sin2 = 1 .

csc2x + sec2x = 1/sin2x + 1/cos2x = (cos2x + sin2x)/(sin2x*cos2x) = 1/(sin2x*cos2x) = csc2x*sec2x

The final expression is the same as the right-hand side of the original equality.

Alan Jun 6, 2014

4 Online Users

avatar