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How do you prove #csc^2xtan^2x-1=tan^2x#?

Answer 1

Daniel Acosta

Prove trig identity

#(1/(sin^2 x))(sin^2 x/cos^2 x) - 1 = #

#= (sin^2 x - sin^2 x.cos^2 x)/(sin^2 x.cos^2 x)# =

#((sin^2 x)(1 - cos^2 x))/(sin^2x.cos^2 x) = #

=#sin^2x/cos^2 x = tan^2 x#

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Answer 2

Quinn Anderson

To prove ( \csc^2(x) \tan^2(x) - 1 = \tan^2(x) ), you can start with the Pythagorean identities for sine and cosine:

[ \sin^2(x) + \cos^2(x) = 1 ]

Then, divide both sides by ( \cos^2(x) ):

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

Using the reciprocal identities for sine and cosine:

[ \csc^2(x) + 1 = \sec^2(x) ]

Rearranging the terms:

[ \csc^2(x) = \sec^2(x) - 1 ]

Now, substitute ( \sec^2(x) = 1 + \tan^2(x) ) (another Pythagorean identity):

[ \csc^2(x) = 1 + \tan^2(x) - 1 ]

[ \csc^2(x) = \tan^2(x) ]

Thus, ( \csc^2(x) \tan^2(x) - 1 = \tan^2(x) ).

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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.