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Which of the following statements is not true? need help asap

Which Of The Following Statements Is Not True Need Help Asap class=

Sagot :

Answer:

B

Step-by-step explanation:

a.Let prove each one.

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

[tex] \frac{1}{ \cos(x) } \frac{ \sin(x) }{ \cos(x) } = [/tex]

[tex] \frac{ \sin(x) ) }{ \cos {}^{2} (x) } [/tex]

[tex] \frac{ \sin(x) }{1 - \sin {}^{2} (x) } [/tex]

Since a is true, that isn't the answer.

b.

[tex] \sec(x) + \cos(x) = \sin(x) \tan(x) [/tex]

[tex] \frac{1}{ \cos(x) } + \cos(x) = \sin(x) \tan(x) [/tex]

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

B doesn't seem right but for the sake of getting better, let see about c and d.

C.

[tex] \csc(x) - \sin(x) = \cot(x) \cos(x) [/tex]

[tex] \frac{1}{ \sin(x) } - \sin(x) = \cot(x) \cos(x) [/tex]

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

[tex] \frac{ \cos {}^{2} (x) }{ \sin(x) } = \cot(x) \cos(x) [/tex]

[tex] \frac{ \cos(x) \cos(x) }{ \sin(x) } = \cot(x) \cos(x) [/tex]

[tex] \frac{ \cos(x) }{ \sin(x) } \times \cos(x) = \cot(x) \cos(x) [/tex]

[tex] \cot(x) \cos(x) = \cot(x) \cos(x) [/tex]

So c is right, so it isn't the answer.

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

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

[tex] \frac{ \cos {}^{2} (x) - \sin {}^{2} (?) }{ \sin(x) \cos(x) } = \cot(x) - \tan(x) [/tex]

[tex] \frac{ \cos {}^{2} (x) }{ \cos(x) \sin(x) } - \frac{ \sin {}^{2} x }{ \cos(x) \sin(x) } [/tex]

[tex] \frac{ \cos(x) }{ \sin(x) } - \frac{ \sin(x) }{ \cos(x) } = \cot(x) - \tan(x) [/tex]

[tex] \cot(x) - \tan(x) = \cot(x) - \tan(x) [/tex]

So D is right, So that isn't isn't answer.

It seems that B is the right answer since it isnt isn't true identity.

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