SOLUTION: Help with verifying a trigonometric identity please?
secx-cosx+cscx-sinx-(sinxtanx)=(cosxcotx)
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Question 591304: Help with verifying a trigonometric identity please?
secx-cosx+cscx-sinx-(sinxtanx)=(cosxcotx)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ -\ \cos(x)\ +\ \csc(x)\ -\ \sin(x)\ -\ \left(\sin(x)\tan(x)\right)\ \equiv^?\ \cos(x)\cot(x))
Use
\ \equiv\ \frac{1}{\cos(\varphi))
\ \equiv\ \frac{1}{\sin(\varphi))
and
\ \equiv\ \frac{\sin(\varphi)}{\cos(\varphi))
to convert everything in the LHS to sine and cosine functions:
}\ -\ \cos(x)\ +\ \frac{1}{sin(x)}\ -\ \sin(x)\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x))
Combine the first and second and third and fourth terms using )
and )
as common denominators:
}{\cos(x)}\ +\ \frac{1\ -\ \sin^2(x)}{sin(x)}\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x))
Apply the Pythagorean Identity:
}{\cos(x)}\ +\ \frac{\cos^2(x)}{sin(x)}\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x))
Note the two additive inverse terms in the LHS and eliminate.
}{sin(x)}\ \equiv^?\ \cos(x)\cot(x))
Decompose the fraction in the LHS:
}{1}\right)\left(\frac{\cos(x)}{sin(x)}\right)\ \equiv^?\ \cos(x)\cot(x))
Finally, use
\ \equiv\ \frac{\cos(\varphi)}{\sin(\varphi))
To write:
\cot(x) \equiv\ \cos(x)\cot(x))
Q.E.D.
John
My calculator said it, I believe it, that settles it
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