How do you show that #sec^2xsin^2x = sec^2x - 1#?
3 Answers
Jul 20, 2017
Use
Jul 20, 2017
Explanation:
#"using the "color(blue)"trigonometric identities"#
#•color(white)(x)sec^2x=tan^2x+1#
#•color(white)(x)secx=1/cosx#
#•color(white)(x)tanx=sinx/cosx#
#"consider the left side"#
#sec^2x-1=(tan^2x+1)-1#
#color(white)(sec^2x-1)=tan^2x#
#color(white)(sec^2x-1)=sin^2x/cos^2x#
#color(white)(sec^2x-1)=1/cos^2x xxsin^2x#
#color(white)(sec^2x-1)=sec^2xsin^2x="right side"#
Jul 20, 2017
As Jim H. has suggested, the fastest way is to use
#sin^2x + cos^2x = 1# #" "bb((1))#
#sec^2x = 1/cos^2x# #" "" "bb((2))#
So:
#color(blue)(sec^2xsin^2x) = sec^2x (1 - cos^2x)#
(use#(1)# )
#= sec^2x - sec^2x cdot cos^2x#
(distribute)
#= sec^2x - 1/cancel(cos^2x) cdot cancel(cos^2x)#
(use#(2)# )
#= color(blue)(sec^2x - 1)#
