See below for the proof of the trigonometry identity [tex]\frac{\tan^2(A)}{\tan(A) - 1} + \frac{\cot(A)}{1 - \tan(A)} =1 + \sec(A)\csc(A)[/tex]
How to prove the trigonometry identity?
The trigonometry equation is given as:
[tex]\frac{\tan^2(A)}{\tan(A) - 1} + \frac{\cot(A)}{1 - \tan(A)} =1 + \sec(A)\csc(A)[/tex]
Rewrite the equation as:
[tex]\frac{\tan^2(A) }{\tan(A) - 1} - \frac{\cot(A)}{ \tan(A) - 1} =1 + \sec(A)\csc(A)[/tex]
Take LCM
[tex]\frac{\tan^2(A) - \cot(A)}{ \tan(A) - 1} =1 + \sec(A)\csc(A)[/tex]
Express cot(A) as 1/tan(A)
[tex]\frac{\tan^2(A) - \frac{1}{\tan(A)}}{ \tan(A) - 1} =1 + \sec(A)\csc(A)[/tex]
Take LCM
[tex]\frac{\tan^3(A) - 1}{\tan(A)(\tan(A) - 1)} =1 + \sec(A)\csc(A)[/tex]
Apply the difference of two cubes on the numerator
[tex]\frac{(\tan(A) - 1)(\tan^2(A) + \tan(A) + 1)}{\tan(A)(\tan(A) - 1)} =1 + \sec(A)\csc(A)[/tex]
Cancel out the common factors
[tex]\frac{\tan^2(A) + \tan(A) + 1}{\tan(A)} =1 + \sec(A)\csc(A)[/tex]
Rewrite as:
[tex]\frac{\tan(A) + \tan^2(A) + 1}{\tan(A)} =1 + \sec(A)\csc(A)[/tex]
Split the fraction
[tex]1 + \frac{\tan^2(A) + 1}{\tan(A)} =1 + \sec(A)\csc(A)[/tex]
Express [tex]\tan^2(A) + 1[/tex] as [tex]\sec^2(A)[/tex]
[tex]1 + \frac{sec^2(A)}{\tan(A)} =1 + \sec(A)\csc(A)[/tex]
Divide [tex]\sec^2(A)[/tex] by [tex]\tan(A)[/tex]
[tex]1 + \frac{1}{\sin(A)\cos(A)} = 1 + \sec(A)\csc(A)[/tex]
Take the inverse of sin and cosine
[tex]1 + \sec(A)\csc(A) = 1 + \sec(A)\csc(A)[/tex]
Both sides of the equation are the same.
Hence, the trigonometry identity [tex]\frac{\tan^2(A)}{\tan(A) - 1} + \frac{\cot(A)}{1 - \tan(A)} =1 + \sec(A)\csc(A)[/tex] has been proved
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