How do I prove that #sin^2x# + #tan^2xsin^2x# = #tan^2x# ?

2 Answers
George C.
Sep 20, 2015

Use #tan x = sin x / cos x# and #sin^2 x + cos^2 x = 1# and rearrange.

Explanation:

By definition:

#tan x = sin x / cos x#

and by Pythagoras:

#sin^2 x + cos^2 x = 1#

So:

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

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

#= sin^2 x + sin^2 x tan^2 x#

İnci
Sep 20, 2015

Using the rules: #tanx = sinx/cosx#

#sin^2x + cos^2x = 1#

Explanation:

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

#sin^2x + sin^4x/(cos^2x) = tan^2x#

#sin^2x + sin^4x/(cos^2x) = sin^2x/(cos^2x)#

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

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

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

#sin^2x = sin^2x#

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