What basic trigonometric identity would you use to verify that csc x sec x cot x = csc^(2)x

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carlosego carlosego · Answer 1 · 2019-02-04T22:30:04+01:00

Answer:

[tex]cotx=\frac{cosx}{sinx}[/tex]

[tex]secx=\frac{1}{cosx}[/tex]

[tex]cscx=\frac{1}{sinx}[/tex]

Step-by-step explanation:

1. Keeping on mind that [tex]secx=\frac{1}{cosx}[/tex] and [tex]cotx=\frac{cos}{sin}[/tex], you can rewrite it as following:

[tex]cscx(\frac{1}{cosx})(\frac{cosx}{sinx})[/tex]

2. Then, when you simplify it, you obtain:

[tex]cscx\frac{1}{sinx}[/tex]

3. So, keeping on mind that [tex]cscx=\frac{1}{sinx}[/tex], you have:

[tex]cscx*cscx=csc^{2}x[/tex]

boffeemadrid boffeemadrid · Answer 2 · 2019-05-31T12:38:42+02:00

Answer:

Step-by-step explanation:

Using the basic trigonometry identity, we have

[tex]cosecx=\frac{1}{sinx}[/tex], [tex]secx=\frac{1}{cosx}[/tex] and [tex]cotx=\frac{cosx}{sinx}[/tex]

Thus, the given equation is:

[tex]{\text}{cosecx secx cotx}=cosec^2x[/tex]

Taking the LHS of the above equation , we get

=[tex]{\text}{cosecx secx cotx}[/tex]

=[tex]cosecx(\frac{1}{cosx})(\frac{cosx}{sinx})[/tex]

=[tex]cosec^2x[/tex]

=RHS

Hence proved.

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