Asked by Sandra
Prove that:
(tan^2x +1)cos2x=2-sec^2x
(tan^2x +1)cos2x=2-sec^2x
Answers
Answered by
oobleck
recalling your basic identities and double-angle formulas, the left side becomes
(sec^2x)(2cos^2x-1)
and you're basically home free...
(sec^2x)(2cos^2x-1)
and you're basically home free...
Answered by
Bosnian
_____________________
Remark:
tan² x = sin² x / cos² x
1 = cos² x / cos² x
cos ( 2 x ) = cos² x - sin² x
sin² x + cos² x = 1
sin² x = 1 - cos² x
_____________________
( tan² x + 1 ) ∙ cos ( 2 x ) = 2 - sec² x
( sin² x / cos² x + cos² x / cos² x ) ∙ ( cos² x - sin² x ) = 2 - sec² x
( sin² x + cos² x ) / cos² x ∙ [ cos² x - ( 1 - cos² x ) ] = 2 - sec² x
1 / cos² x ∙ ( cos² x - 1 + cos² x ) = 2 - sec² x
1 / cos² x ∙ ( 2 cos² x - 1 ) = 2 - sec² x
1 / cos² x ∙ 2 cos² x + 1 / cos² x ∙ ( - 1 ) = 2 - sec² x
2 cos² x / cos² x - 1 / cos² x = 2 - sec² x
2 - 1 / cos² x = 2 - sec² x
2 - sec² x = 2 - sec² x
Remark:
tan² x = sin² x / cos² x
1 = cos² x / cos² x
cos ( 2 x ) = cos² x - sin² x
sin² x + cos² x = 1
sin² x = 1 - cos² x
_____________________
( tan² x + 1 ) ∙ cos ( 2 x ) = 2 - sec² x
( sin² x / cos² x + cos² x / cos² x ) ∙ ( cos² x - sin² x ) = 2 - sec² x
( sin² x + cos² x ) / cos² x ∙ [ cos² x - ( 1 - cos² x ) ] = 2 - sec² x
1 / cos² x ∙ ( cos² x - 1 + cos² x ) = 2 - sec² x
1 / cos² x ∙ ( 2 cos² x - 1 ) = 2 - sec² x
1 / cos² x ∙ 2 cos² x + 1 / cos² x ∙ ( - 1 ) = 2 - sec² x
2 cos² x / cos² x - 1 / cos² x = 2 - sec² x
2 - 1 / cos² x = 2 - sec² x
2 - sec² x = 2 - sec² x
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.