Question #4352a

3 Answers
Mar 26, 2017

Answer:

#1#

Explanation:

If #3sin^2x-sin^4x=1# then #sin^2x=(3-sqrt(5))/2#

and

#cos^2x=1-sin^2x=1-(3-sqrt(5))/2#

then

#tan^2x+tan^4x =((3-sqrt(5))/2)/(1-(3-sqrt(5))/2)+(((3-sqrt(5))/2)/(1-(3-sqrt(5))/2))^2=1#

Mar 27, 2017

Answer:

#3 - sin^2 x#

Explanation:

Develop the left side of the equation:
#3sin^2 x - sin^4 x = 1#
#sin^2 x(3 - sin^2 x) = 1#
From this equation we get:
#1/sin^2 x = 3 - sin^2 x# (1)

Develop the right side:
#tan^2 x + tan^4 x = tan^2 x(1 + tan^2 x) = tan^2 x(1/cos^2 x) =#
#= (cos^2 x)/(sin^2 x)(1/(cos^2 x)) = 1/(sin^2 x) (2)#
Compare (1) and (2) -->
#tan^2 x + tan^4 x = 3 - sin^2 x#

Mar 27, 2017

#3sin^2x-sin^4x=1#

Dividing bothsides by #cos^4x#

#=>(3sin^2x)/cos^4x-sin^4x/cos^4x=1/cos^4x#

#=>3tan^2xsec^2x-tan^4x=sec^4x#

#=>3tan^2x(1+tan^2x)-tan^4x=(1+tan^2x)^2#

#=>3tan^2x+3tan^4x-tan^4x=1+2tan^2x+tan^4x#

#=>3tan^2x-2tan^2x+3tan^4x-tan^4x-tan^4x=1#

#=>tan^2x+tan^4x=1#