How do you verify #tan^2x-tan^2x*sin^2x=sin^2x#?

2 Answers
Shwetank Mauria
Apr 8, 2018

Please see below.

Explanation:

#tan^2x-tan^2x*sin^2x#

= #tan^2x(1-sin^2x)#

= #tan^2xcos^2x#

= #sin^2x/cos^2x*cos^2x#

= #sin^2x#

Jim G.
Apr 8, 2018

#"see explanation"#

Explanation:

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

#•color(white)(x)tanx=sinx/cosx#

#•color(white)(x)sin^2x+cos^2x=1#

#"consider the left side"#

#"take out a common factor "tan^2x#

#tan^2x(1-sin^2x)#

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

#=sin^2x=" right side "rArr"verified"#

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