Chapter 8.5, Problem 25WE

a.

To determine

To find: The value of $\tan {35}^{\circ }+\tan {25}^{\circ }$ .

Answer to Problem 25WE

  $\tan {35}^{\circ }+\tan {25}^{\circ }=1.1665$

Explanation of Solution

Given:

  $\tan {35}^{\circ }+\tan {25}^{\circ }$

Calculation:

  $\tan {35}^{\circ }+\tan {25}^{\circ }\approx 0.7002+0.4663=1.1665$

Conclusion:

Therefore, $\tan {35}^{\circ }+\tan {25}^{\circ }\approx 0.7002+0.4663=1.1665$

b.

To determine

To find the value of $\tan \left({35}^{\circ }+{25}^{\circ }\right)$ .

Answer to Problem 25WE

  $\sqrt{3}=1.7320$

Explanation of Solution

Given:

  $\tan \left({35}^{\circ }+{25}^{\circ }\right)$

Formula used:

  $\tan {60}^{\circ }=\sqrt{3}$

Calculation:

  $\begin{array}{l}\Rightarrow \tan \left({35}^{\circ }+{25}^{\circ }\right)\\ \Rightarrow \tan {60}^{\circ }\approx \sqrt{3}=1.7320\end{array}$

Conclusion:

Therefore, $\tan \left({35}^{\circ }+{25}^{\circ }\right)\approx \sqrt{3}=1.7320$ .

c.

To determine

To show: $\tan A+\tan B\ne \tan \left(A+B\right)$ .

Answer to Problem 25WE

No

Explanation of Solution

Given:

  $\tan A+\tan B$

Formula used:

Trigonometric formula:

  $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Calculation:

Since,

  $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

  $\Rightarrow \tan A+\tan B=\tan \left(A+B\right)\left(1-\tan A\tan B\right)$

  $\ne \tan \left(A+B\right)$

Conclusion:

Therefore, $\tan A+\tan B\ne \tan \left(A+B\right)$ .