How do you prove # sin^2x + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x#?

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Answer 1 · 2018-03-28T13:18:57

Please refer to the Explanation.

#sin^2x+tan^2x/(1-tan^2x)#,

#=(1-cos^2x)+tan^2x/(1-tan^2x)#,

#=1+(tan^2x/(1-tan^2x))-cos^2x#,

#={(1-tan^2x)+tan^2x}/(1-tan^2x)-cos^2x#,

#=1/(1-tan^2x)-cos^2x#,

#={1-:(1-sin^2x/cos^2x)}-cos^2x#,

#={1-:(cos^2x-sin^2x)/cos^2x}-cos^2x#,

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

Please check the Problem.

Answer 2 · 2018-03-28T13:26:24

Please see below.
# sin^2x +(color(red)(2)tan^2x)/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x# ?

#sin^2x+(color(red)(2)tan^2x)/(1-tan^2x)=1/(cos^2x-sin^2x)-cos^2x#

#LHS=sin^2x+(2tan^2x)/(1-tan^2x)#

#=1-cos^2x+(2tan^2x)/(1-tan^2x)#

#=[1+(2tan^2x)/(1-tan^2x)]-cos^2x#

#=(1-tan^2x+2tan^2x)/(1-tan^2x)-cos^2x#

#=(1+tan^2x)/(1-tan^2x)-cos^2x#

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

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

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

#=RHS#