SOLUTION: I do not get how to prove that left side is equal to right side: tan<sup>2</sup>@ - sin<sup>2</sup>@ = sin<sup>2</sup>@*tan<sup>2</sup>@ </pre>

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Question 194509: I do not get how to prove that left side is equal to right side:
tan2@ - sin2@ = sin2@*tan2@
Found 2 solutions by RAY100, Edwin McCravy:
Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
tan^2a -sin^2a = sin^2a * tan^2a,,,,let a = theta
,
but tan = sin / cos
,
sin^2 a/cos^2a -sin^2a = sin^2a *tan^2a
,,
remember cos^2 = sin^2 =1,,, cos^2a = 1=sin^2a
,
sin^2a / ( 1-sin^2a) - sin^2a = sin^2a * tan^2a
,
using lcd = (1-sin^2a)
,
( sin^2a -sin^2a (1-sin^2a) )/ (1-sin^2a) = sin^2a *tan^2a
,
( sin^2a -sin^2a +sin4a ) (1-sin^2a) = sin^2a * tan^2a
,
but cos^2a = 1 - sin^2a
,
sin^4a / cos^2a = sin^2a * tan^2a
,
with tan = sin / cos
,
sin^2a * tan^2a = sin^2a * tan^2a

Answer by Edwin McCravy(20066) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution:

tan2@ - sin2@  =   sin2@*tan2@

sin2@   sin2@
----- - -----
cos2@     1

sin2@   sin2@*cos2@
----- - -----------
cos2@       1*cos2@

sin2@ - sin2@*cos2@
-------------------
       cos2@

 sin2@(1 - cos2@)
------------------
      cos2@

    sin2@(sin2@)
  ---------------
       cos2@

  /sin2@\  
 | ------| (sin2@)
  \cos2@/


    tan2@*sin2@

    sin2@*tan2@

Edwin