## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 227: 30

#### Answer

$$\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}=-\sqrt3$$

#### Work Step by Step

$$X=\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}$$ From the identity of the sum of tangent: $$\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)$$ So here $X$ actually follows the above identity with $A=\frac{5\pi}{12}$ and $B=\frac{\pi}{4}$. Therefore, $X$ can also be rewritten as $$X=\tan\Big(\frac{5\pi}{12}+\frac{\pi}{4}\Big)$$ $$X=\tan\frac{5\pi+3\pi}{12}$$ $$X=\tan\frac{8\pi}{12}=\tan\frac{2\pi}{3}$$ From Section 5.3, as cofunction identities: $$\tan\theta=\cot\Big(\frac{\pi}{2}-\theta\Big)$$ Therefore, $$X=\cot\Big(\frac{\pi}{2}-\frac{2\pi}{3}\Big)$$ $$X=\cot\Big(-\frac{\pi}{6}\Big)$$ Also, we know $$\cot(-\theta)=-\cot\theta$$ $$X=-\cot\Big(\frac{\pi}{6}\Big)$$ $$X=-\sqrt3$$ which means $$\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}=-\sqrt3$$

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