Chapter 7 - 7.4 - Sum and Difference Equations - 7.4 Exercises - Page 539: 62

The identity is verified. $tan(\frac{\pi}{4}-θ)=\frac{1-tan~θ}{1+tan~θ}$

Work Step by Step

$tan(u-v)=\frac{tan~u-tan~v}{1+tan~u~tan~v}$ $tan(\frac{\pi}{4}-θ)=\frac{tan~\frac{\pi}{4}-tan~θ}{1+tan~\frac{\pi}{4}~tan~θ}=\frac{1-tan~θ}{1+(1)tan~θ}=\frac{1-tan~θ}{1+tan~θ}$

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