Câu 25:

$1 + \tan^2 15^\circ = \frac{1}{\cos^2 15^\circ} = \frac{4}{2+\sqrt{3}} = 4(2-\sqrt{3}) = 8 - 4\sqrt{3}$

Suy ra: $\tan^2 15^\circ = 7 - 4\sqrt{3} = (2-\sqrt{3})^2$

Vì $0^\circ < 15^\circ < 90^\circ$ nên $\tan 15^\circ > 0$

Vậy $\tan 15^\circ = 2 - \sqrt{3}$

Đáp án: C. $2 - \sqrt{3}$

Câu 26:

Vì $\frac{\pi}{2}$ < $\alpha$ < $\pi$ nên $\sin \alpha > 0$

Ta có: $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{4}{25} = \frac{21}{25}$

Suy ra: $\sin \alpha = \frac{\sqrt{21}}{5}$

Vậy $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = - \frac{\sqrt{21}}{2}$

Đáp án: A. $- \frac{\sqrt{21}}{2}$

Câu 27:

Vì $\pi$ < $\alpha$ < $\frac{3\pi}{2}$ nên $\cos \alpha < 0$

Ta có: $1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}$ => $\cos^2 \alpha = \frac{1}{(1+\tan^2 \alpha)} = \frac{1}{6}$

Suy ra: $\cos \alpha = - \frac{\sqrt{6}}{6}$ (vì $\cos \alpha < 0$)

Đáp án: C. $- \frac{\sqrt{6}}{6}$

Câu 28:

Vì $90^\circ < \alpha < 180^\circ$ nên $\cos \alpha < 0$

Ta có: $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25}$

Suy ra: $\cos \alpha = - \frac{4}{5}$ (vì $\cos \alpha < 0$)

Vậy $\cot \alpha = \frac{\cos \alpha}{\sin \alpha} = - \frac{4}{3}$

Đáp án: D. $\cot \alpha = - \frac{4}{3}$

Câu 29:

Vì $\sin \alpha > 0$ và $\cos \alpha < 0$ nên góc $\alpha$ thuộc góc phần tư thứ II.

Trong góc phần tư thứ II, $\tan \alpha < 0$

Ta có: $\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{(\sin^2 \alpha)}{(1 - \sin^2 \alpha)} = \frac{4}{5}$

Suy ra: $\tan \alpha = - \frac{2}{\sqrt{5}}$ = $- \frac{2\sqrt{5}}{5}$ (vì $\tan \alpha < 0$)

Đáp án: A. $- \frac{2\sqrt{5}}{5}$

Vì $\frac{\pi}{2}$ < $\alpha$ < $\pi$ nên $\cos \alpha < 0$.

Ta có: $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{1}{9} = \frac{8}{9}$

Suy ra: $\cos \alpha = - \frac{2\sqrt{2}}{3}$ (vì $\cos \alpha < 0$)

Đáp án: B. $\cos \alpha = - \frac{2\sqrt{2}}{3}$

Câu 31:

Ta có: $\tan \frac{\alpha}{2}$ + $\cot \frac{\alpha}{2}$ = $\frac{\sin(\alpha/2)}{\cos(\alpha/2)}$ + $\frac{\cos(\alpha/2)}{\sin(\alpha/2)}$

= $\frac{\sin^2(\alpha/2) + \cos^2(\alpha/2)}{\sin(\alpha/2).\cos(\alpha/2)}$ = $\frac{2}{\sin \alpha}$

Vì $\frac{\pi}{2}$ < $\alpha$ < $\pi$ nên $\sin \alpha > 0$.

Ta có: $1 + \cot^2 \alpha = \frac{1}{\sin^2 \alpha}$ => $\sin^2 \alpha = \frac{1}{(1 + \cot^2 \alpha)} = \frac{1}{19}$

Suy ra: $\sin \alpha = \frac{1}{\sqrt{19}}$ (vì $\sin \alpha > 0$)

Vậy $\tan \frac{\alpha}{2}$ + $\cot \frac{\alpha}{2}$ = $2\sqrt{19}$

Đáp án: D. $\sqrt{19}$

Câu 32:

Ta có: $(\sin \alpha + \cos \alpha)^2 = \frac{9}{4}$

<=> $\sin^2 \alpha + 2\sin \alpha.\cos \alpha + \cos^2 \alpha = \frac{9}{4}$

<=> $1 + \sin 2\alpha = \frac{9}{4}$

<=> $\sin 2\alpha = \frac{5}{4}$

Đáp án: A. $\frac{5}{4}$