SOLUTION: I would love love love help proving the following identity:
1 + tan(theta)/ 1 + cot(theta) = 1- tan(theta)/cot(theta) - 1
thank you much
Algebra.Com
Question 645659: I would love love love help proving the following identity:
1 + tan(theta)/ 1 + cot(theta) = 1- tan(theta)/cot(theta) - 1
thank you much
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I should make you repost this because your rendering of the equation is so ambiguous; there are several ways to interpret what does or does not go into each numerator and denominator. But since my first guess at what you meant works out, I'll give you a break this time. Next time use sufficient parentheses so that it is clear what you mean. I'm interpreting your input as:
(1 + tan(t))/( 1 + cot(t)) = (1 - tan(t))/(cot(t) - 1)
BTW, there is no need to spell out theta. You can use any variable since it all comes out in the wash at the end anyway.
}{1\ +\ \cot(\theta)}\ \equiv^?\ \frac{1\ -\ \tan(\theta)}{\cot(\theta)\ -\ 1})
Use the identities for tangent and cotangent in terms of sine and cosine:
}{\cos(\theta)}}{1\ +\ \frac{\cos(\theta)}{\sin(\theta)}}\ \equiv^?\ \frac{1\ -\ \frac{\sin(\theta)}{\cos(\theta)}}{\frac{\cos(\theta)}{\sin(\theta)}\ -\ 1})
Apply the LCD for each numerator and denominator expression:
\ +\ \sin(\theta)}{\cos(\theta)}}{\frac{\sin(\theta)\ +\ \cos(\theta)}{\sin(\theta)}}\ \equiv^?\ \frac{\frac{\cos(\theta)\ -\ \sin(\theta)}{\cos(\theta)}}{\frac{\cos(\theta)\ -\ \sin(\theta)}{\sin(\theta)}})
Next, invert and multiply to divide on each side. You can write it out for yourself if you want to verify, but note that on each side of the equation the numerator of the numerator is equal to the numerator of the denominator, hence, when you invert and multiply on both sides you are left with:
}{\cos(\theta)}\ \equiv\ \frac{\sin(\theta)}{\cos(\theta)})
John
My calculator said it, I believe it, that settles it
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