Trigonometric identities are used to prove or disprove trigonometric ratios
The trigonometry identity is given as:
[tex]\frac{\cos(\pi + x) \cos(-x)}{\sin(\pi - x) \cos(\frac{\pi}{2} + x)} = \cot^2x[/tex]
Simplify the trigonometry identity, as follows:
[tex]\frac{-\cos(x) \cos(x)}{\sin( x) \cdot - \sin( x)} = \cot^2x[/tex]
This is so because:
[tex]\cos(\pi + x) = -\cos(x)[/tex]
[tex]\cos(-x) = \cos(x)[/tex]
[tex]\sin(\pi -x) = \sin(x)[/tex]
[tex]\cos(\frac{\pi}{2} + x)} = -\sin(x)[/tex]
So, we have:
[tex]\frac{-\cos(x) \cos(x)}{\sin( x) \cdot - \sin( x)} = \cot^2x[/tex]
Evaluate the products
[tex]\frac{-\cos^2(x)}{-\sin^2(x)} = \cot^2x[/tex]
Cancel out the common factors
[tex]\frac{\cos^2(x)}{\sin^2(x)} = \cot^2x[/tex]
Rewrite the trigonometry identity, as follows:
[tex](\frac{\cos(x)}{\sin(x)})^2 = \cot^2x[/tex]
In trigonometry,
[tex]\frac{\cos(x)}{\sin(x)} = \cot x[/tex]
So, we have:
[tex](\cot(x))^2 = \cot^2x[/tex]
Remove the bracket
[tex]\cot^2x = \cot^2x[/tex]
The above equation is true.
Hence, [tex]\frac{\cos(\pi + x) \cos(-x)}{\sin(\pi - x) \cos(\frac{\pi}{2} + x)} = \cot^2x[/tex] has been proved
Read more about trigonometric identities at:
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