Prove the identity tan^2x-sin^2x is same as (tan^2x)(sin^2x)?

Prove the identity #tan^2x-sin^2x #is same as
#(tan^2x)(sin^2x)?#

2 Answers
Dillin N.
Dec 7, 2017

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

Explanation:

Assuming #tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)#, start off by rewriting #tan^2(x)# in to its #sin(x)# and #cos(x)# components.

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

Next find a common denominator (LCD: #cos^2(x)*1#)

#sin^2(x)/cos^2(x)*(1/1) - sin^2(x)*cos^2(x)/cos^2(x) rarr sin^2(x)/cos^2(x) - (sin^2(x)cos^2(x))/cos^2(x)#

Combine in to a single fraction and factor out a #sin^2(x)#.

#(sin^2(x)-sin^2(x)cos^2(x))/cos^2(x) rarr sin^2(x)*sin^2(x)/cos^2(x)#

Finally just rewrite

#sin^2(x)*sin^2(x)/cos^2(x) rarr sin^2(x)tan^2(x)#

Ratnaker Mehta
Dec 7, 2017

Please refer to a Proof given in the Explanation.

Explanation:

We have,

#tan^2x-sin^2x,#

#=sin^2x/cos^2x-sin^2x,#

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

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

#=sin^2x{sin^2x/cos^2x},#

#=sin^2x*tan^2x.#

Hence, the Proof.

Related questions
Impact of this question
67604 views around the world
You can reuse this answer
Creative Commons License