Prove that:

Question

(tan^2x +1)cos2x=2-sec^2x

Answer 1

recalling your basic identities and double-angle formulas, the left side becomes
(sec^2x)(2cos^2x-1)

and you're basically home free...

Answer 2

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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
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( 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