Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 43

Answer

The required value of $\tan 75{}^\circ $ is $2+\sqrt{3}$.

Work Step by Step

We have to find the value of $\tan 75{}^\circ $; the formula is used which represents the tan identity in term of cos and sine. The term $\tan 75{}^\circ $ comes under the first quadrant where the value of all the trigonometric functions is positive. And the value of the first quadrant angle is between $0{}^\circ \,\text{to}\,9\text{0}{}^\circ $. $\begin{align} & \tan 75{}^\circ =\tan \frac{150{}^\circ }{2} \\ & =\frac{1-\cos 150{}^\circ }{\sin 150{}^\circ } \\ & =\frac{1-\left( -\frac{\sqrt{3}}{2} \right)}{\frac{1}{2}} \\ & =2+\frac{\sqrt{3}}{2}\times 2 \end{align}$ So, in the aforementioned way the equation is solved: $\begin{align} & \tan 75{}^\circ =2+\frac{\sqrt{3}}{{}} \\ & =2+\sqrt{3} \end{align}$ Thus, the value of $\tan 75{}^\circ $ is $2+\sqrt{3}$.
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