How do you verify #1-cos(2x)*sec^2(x) = (tan(x))^2#?

Answer 1

LHS=#1-cos(2x)*sec^2(x) # putting #cos2x=2cos^2x-1# #=1-(2cos^2x-1)*sec^2(x) # #=1-(2cos^2x*sec^2x-sec^2(x)) # #=1-2+sec^2x=sec^2x-1=tan^2x=(tanx)^2=RHS#

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Answer 2

Samantha Abate

To verify the equation (1 - \cos(2x) \cdot \sec^2(x) = \tan^2(x)), we can start by expressing (\sec^2(x)) and (\cos(2x)) in terms of (\tan(x)).

Recall that (\sec^2(x) = \frac{1}{\cos^2(x)}) and (\cos(2x) = \cos^2(x) - \sin^2(x)).

Substituting these expressions into the equation gives:

[1 - (\cos^2(x) - \sin^2(x)) \cdot \frac{1}{\cos^2(x)} = \tan^2(x)]

Simplify the expression by distributing and combining like terms:

[1 - \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)} = \tan^2(x)]

[1 - \frac{\cos^2(x)}{\cos^2(x)} + \frac{\sin^2(x)}{\cos^2(x)} = \tan^2(x)]

[1 - 1 + \frac{\sin^2(x)}{\cos^2(x)} = \tan^2(x)]

[ \frac{\sin^2(x)}{\cos^2(x)} = \tan^2(x) ]

Since (\tan^2(x) = \tan^2(x)), the equation is verified.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.