How do you verify #(tan^2x + 1) / tan^2x = csc^2x#?

2 Answers
Jun 19, 2016

Answer:

Use the Pythagorean identity #tan^2x + 1 = sec^2x# to start the simplification on the left side.

Explanation:

#sec^2x/tan^2x =#

Recall that #sec^2x = 1/cos^2x#, and that #tan^2x = sin^2x/cos^2x#

#(1/cos^2x)/(sin^2x/(cos^2x)) =#

#1/cos^2x xx cos^2x/sin^2x =#

#1/sin^2x =#

Remember that #1/sinx = cscx#

#csc^2x =#

Identity proved!!!

Hopefully this helps!

Jun 19, 2016

The definition of tangent is such that

#tan(x)=sin(x)/cos(x)#, then the left side of the equation is

#((sin(x)^2/cos(x)^2)+1)/(sin(x)^2/cos(x)^2)#

#=((sin(x)^2/cos(x)^2)+1)*cos(x)^2/sin(x)^2#

#=1+cos(x)^2/sin(x)^2#

#=(sin(x)^2+cos(x)^2)/sin(x)^2#

and, using the fundamental property of #sin# and #cos# that #sin(x)^2+cos(x)^2=1# we have

#(sin(x)^2+cos(x)^2)/(sin(x)^2) = 1/(sin(x)^2)=csc(x)^2#.