As shown in the figure, in triangle ABC with $\overline { \mathrm { AB } } = 5 , \overline { \mathrm { AC } } = 2 \sqrt { 5 }$, let D be the foot of the perpendicular from vertex A to segment BC.
For point E that divides segment AD internally in the ratio $3 : 1$, we have $\overline { \mathrm { EC } } = \sqrt { 5 }$. If $\angle \mathrm { ABD } = \alpha , \angle \mathrm { DCE } = \beta$, what is the value of $\cos ( \alpha - \beta )$? [4 points]
(1) $\frac { \sqrt { 5 } } { 5 }$
(2) $\frac { \sqrt { 5 } } { 4 }$
(3) $\frac { 3 \sqrt { 5 } } { 10 }$
(4) $\frac { 7 \sqrt { 5 } } { 20 }$
(5) $\frac { 2 \sqrt { 5 } } { 5 }$

In an isosceles triangle ABC with $\overline { \mathrm { AB } } = \overline { \mathrm { AC } }$, let $\angle \mathrm { A } = \alpha , \angle \mathrm { B } = \beta$. If $\tan ( \alpha + \beta ) = - \frac { 3 } { 2 }$, what is the value of $\tan \alpha$? [3 points]
(1) $\frac { 5 } { 2 }$
(2) $\frac { 12 } { 5 }$
(3) $\frac { 23 } { 10 }$
(4) $\frac { 11 } { 5 }$
(5) $\frac { 21 } { 10 }$

If $\tan a = \frac { 1 } { 3 } , \tan ( a + b ) = \frac { 1 } { 2 }$, then $\tan b =$
(A) $\frac { 1 } { 7 }$
(B) $\frac { 1 } { 6 }$
(C) $\frac { 5 } { 7 }$
(D) $\frac { 5 } { 6 }$

8. Given $\tan \alpha = - 2 , \tan ( \alpha + \beta ) = \frac { 1 } { 7 }$, then the value of $\tan \beta$ is $\_\_\_\_$ .

Given $\tan \left( \alpha - \frac { 5 \pi } { 4 } \right) = \frac { 1 } { 5 }$, then $\tan \alpha = $ \_\_\_\_ .

14. Given $\tan \left( \alpha + \frac { \pi } { 4 } \right) = 6$, then $\tan \alpha = $ \_\_\_\_.

Given $2 \tan \theta - \tan \left( \theta + \frac { \pi } { 4 } \right) = 7$ , then $\tan \theta =$
A. $- 2$
B. $- 1$
C. $1$
D. $2$

Given that $\alpha$ is an angle in the first quadrant, $\beta$ is an angle in the third quadrant, $\tan \alpha + \tan \beta = 4$, $\tan \alpha \tan \beta = \sqrt { 2 } + 1$, then $\sin ( \alpha + \beta ) =$ $\_\_\_\_$ .

Given $\cos ( \alpha + \beta ) = m , \tan \alpha \tan \beta = 2$ , then $\cos ( \alpha - \beta ) =$
A. $- 3 m$
B. $- \frac { m } { 3 }$
C. $\frac { m } { 3 }$
D. $3 m$

110. If the terminal side of arc $\alpha$ is in the second quadrant and $\sin\alpha = \dfrac{\sqrt{2}}{10}$, what is the value of $\cos\!\left(\dfrac{11\pi}{4} + \alpha\right)$?
$$-\frac{4}{5} \quad (1) \qquad -\frac{3}{5} \quad (2) \qquad \frac{3}{5} \quad (3) \qquad \frac{4}{5} \quad (4)$$

33. If $a + \beta = \pi / 2$ and $\beta + \gamma = a$, then tan $a$ equals:
(A) $2 ( \tan \beta + \tan \gamma )$
(B) $\tan \beta + \tan \gamma$
(C) $\tan \beta + 2 \tan \gamma$
(D) $2 \tan \beta + \tan \gamma$

Considering only the principal values of the inverse trigonometric functions, the value of
$$\tan \left( \sin ^ { - 1 } \left( \frac { 3 } { 5 } \right) - 2 \cos ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right) \right)$$
is
(A) $\frac { 7 } { 24 }$
(B) $\frac { - 7 } { 24 }$
(C) $\frac { - 5 } { 24 }$
(D) $\frac { 5 } { 24 }$

If $\cos(\alpha + \beta) = \frac{3}{5}$, $\sin(\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$, then $\tan 2\alpha$ is equal to:
(1) $\frac{21}{16}$
(2) $\frac{63}{52}$
(3) $\frac{33}{52}$
(4) $\frac{63}{16}$

If $\alpha = \cos^{-1}\frac{3}{5}$, $\beta = \tan^{-1}\frac{1}{3}$, where $0 < \alpha, \beta < \frac{\pi}{2}$, then $\alpha - \beta$ is equal to
(1) $\tan^{-1}\frac{9}{14}$
(2) $\cos^{-1}\frac{9}{5\sqrt{10}}$
(3) $\sin^{-1}\frac{9}{5\sqrt{10}}$
(4) $\tan^{-1}\frac{9}{5\sqrt{10}}$

$2 \pi - \left( \sin ^ { - 1 } \frac { 4 } { 5 } + \sin ^ { - 1 } \frac { 5 } { 13 } + \sin ^ { - 1 } \frac { 16 } { 65 } \right)$ is equal to:
(1) $\frac { \pi } { 2 }$
(2) $\frac { 5 \pi } { 4 }$
(3) $\frac { 3 \pi } { 2 }$
(4) $\frac { 7 \pi } { 4 }$

If $S$ is the sum of the first 10 terms of the series, $\tan ^ { - 1 } \left( \frac { 1 } { 3 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 13 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 21 } \right) + \ldots\ldots$. then $\tan ( S )$ is equal to :
(1) $\frac { 5 } { 6 }$
(2) $\frac { 5 } { 11 }$
(3) $- \frac { 5 } { 6 }$
(4) $\frac { 10 } { 11 }$

For $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$ let $3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$ and a real number $k$ be such that $\tan\alpha = k\tan\beta$. Then the value of $k$ is equal to
(1) $-5$
(2) $5$
(3) $\frac{2}{3}$
(4) $-\frac{2}{3}$

If $\frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 4 }$, then $\cos ^ { - 1 } \left( \frac { 12 } { 13 } \cos x + \frac { 5 } { 13 } \sin x \right)$ is equal to
(1) $x - \tan ^ { - 1 } \frac { 4 } { 3 }$
(2) $x + \tan ^ { - 1 } \frac { 4 } { 5 }$
(3) $x - \tan ^ { - 1 } \frac { 5 } { 12 }$
(4) $x + \tan ^ { - 1 } \frac { 5 } { 12 }$

$\cos \left( \sin ^ { - 1 } \frac { 3 } { 5 } + \sin ^ { - 1 } \frac { 5 } { 13 } + \sin ^ { - 1 } \frac { 33 } { 65 } \right)$ is equal to:
(1) 1
(2) 0
(3) $\frac { 32 } { 65 }$
(4) $\frac { 33 } { 65 }$

If $\cot x = \frac { 5 } { 12 }$ for some $\mathrm { x } \in \left( \pi , \frac { 3 \pi } { 2 } \right)$ then $\sin 7 x \left( \cos \frac { 13 x } { 2 } + \sin \frac { 13 x } { 2 } \right) + \cos 7 x \left( \cos \frac { 13 x } { 2 } - \sin \frac { 13 x } { 2 } \right)$ is equal to