Question

Precalculus please help me right answers only

Answer 1

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.