Can someone show me !

Question

0

870

1

+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

sportslover Jun 6, 2014

+33661

+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

Alan · Answer 1 · 2014-06-06T08:58:21

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.