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Alchen [17]
3 years ago
10

Precalculus please help me right answers only

Mathematics
1 answer:
RSB [31]3 years ago
6 0

Consider all options.

A.

\sec^2x-\csc^2x=1.

Note that

\sec x=\dfrac{1}{\cos x},\\ \\\csc x=\dfrac{1}{\sin x}.

Then

\sec^2x-\csc^2x=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}=\dfrac{-4\cos (2x)}{\sin^2(2x)}=\\ \\=-4\dfrac{\cot (2x)}{\sin (2x)}\neq 1.

This option is false.

B.

Since \cos (2x)=\cos^2x-\sin^2x, then  \sin^2x-\cos^2 x=-\cos (2x)\neq 1.

This option is false.

C.

Consider

1+\cot^2x=1+\dfrac{\cos^2x}{\sin^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x}=\dfrac{1}{\sin^2x}=\csc^2x.

This option is true.

D.

\sec^2x-\tan^2x=\dfrac{1}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1.

This option is true.

Answer: False A and B.

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Hope that helped : )

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