How do you simplify #(sec^2x-1)/sin^2x#?

3 Answers
Apr 13, 2018

#(sec^2(x)-1)/sin^2(x)=sec^2(x)#

Explanation:

First, convert all of the trigonometric functions to #sin(x)# and #cos(x)#:

#(sec^2(x)-1)/sin^2(x)#

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

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

Use the identity #sin^2(x)+cos^2(x)=1#:

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

Canceling out the #sin^2(x)# present in both the numerator and the denominator:

#=1/cos^2(x)#

#=sec^2(x)#

Apr 13, 2018

The answer is #sec^2x#.

Explanation:

We know that,
#sec^2x-1=tan^2x#

Therefore,#(sec^2x-1)/sin^2x#
=#tan^2x/sin^2x#
=#sin^2x/cos^2x*1/sin^2x#
=#1/cos^2x#

=#sec^2x#

Apr 13, 2018

#sec^2x#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)secx=1/cosx#

#•color(white)(x)sin^2x+cos^2x=1#

#rArr(1/cos^2x-cos^2x/cos^2x)/sin^2x#

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

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

#=cancel(sin^2x)/cos^2x xx1/cancel(sin^2x)#

#=1/cos^2x=sec^2x#