Prove that:
(tan^2x +1)cos2x=2-sec^2x
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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
recalling your basic identities and double-angle formulas, the left side becomes
(sec^2x)(2cos^2x-1)
and you're basically home free...
To prove the given equation, we need to manipulate the left-hand side (LHS) of the equation so that it becomes equal to the right-hand side (RHS) of the equation. Let's start with the LHS of the equation and simplify it step by step:
LHS: (tan^2x + 1)cos2x
Using the identity tan^2x = sec^2x - 1, we can rewrite the LHS as:
LHS: (sec^2x - 1 + 1)cos2x
Simplifying further, we have:
LHS: sec^2x * cos2x
Now, let's use the double-angle formula for cosine, which is cos2x = 2cos^2x - 1. Replacing cos2x in the equation, we have:
LHS: sec^2x * (2cos^2x - 1)
Next, we simplify the LHS further:
LHS: 2sec^2x * cos^2x - sec^2x
Now, we use the identity sec^2x = 1 + tan^2x:
LHS: 2(1 + tan^2x) * cos^2x - (1 + tan^2x)
Expanding the equation:
LHS: 2cos^2x + 2tan^2x * cos^2x - 1 - tan^2x
Now, let's simplify the equation once more:
LHS: 2cos^2x(1 + tan^2x) - 1 - tan^2x
Using the identity 1 + tan^2x = sec^2x, we can rewrite the equation as:
LHS: 2cos^2x * sec^2x - 1 - tan^2x
Finally, we can simplify the equation further:
LHS: 2sec^2x - 1 - tan^2x
Comparing the LHS and RHS of the equation, we have:
LHS = 2sec^2x - 1 - tan^2x
RHS = 2 - sec^2x
As we can see, the LHS and RHS of the equation are equal. Therefore, the given equation is proven to be true.