Bài tập 5. Chứng minh rằng với mọi góc $\alpha$ ($0^{\circ} \leq \alpha \leq 180^{\circ}$), ta đều có:
a. $cos^{2}\alpha$ + $sin^{2}\alpha$ = 1
b. tan$\alpha$. cot$\alpha$ = 1 ($0^{\circ} < \alpha < 180^{\circ}$, $\alpha \neq 90^{\circ}$)
c. 1 + $tan^{2}\alpha$ = $\frac{1}{cos^{2}\alpha}$
d. 1 + $cot^{2}\alpha$ = $\frac{1}{sin^{2}\alpha}$ ($0^{\circ} < \alpha < 180^{\circ}$, $\alpha \neq 90^{\circ}$)
Bài Làm:
a.
Xét tam giác ABC vuông tại A, ta có:
sinB = $\frac{AC}{BC}$; cosB = $\frac{AB}{BC}$
$cos^{2}B$ + $sin^{2}B$ = $\frac{AB}{BC}^{2}$ + $\frac{AC}{BC}^{2}$ = $\frac{AC^{2} + AB^{2}}{BC^{2}}$ = $\frac{BC^{2}}{BC^{2}}$ = 1 (theo định lí Pytago: $AB^{2} + AC^{2} = BC^{2}$)
Vậy $cos^{2}\alpha$ + $sin^{2}\alpha$ = 1
b. Ta có: tan$\alpha$ = $\frac{sin\alpha}{cos\alpha}$; cot$\alpha$ = $\frac{cos\alpha}{sin\alpha}$
$\Rightarrow$ tan$\alpha$. cot$\alpha$ = $\frac{sin\alpha}{cos\alpha}$. $\frac{cos\alpha}{sin\alpha}$ = 1 (đpcm)
c. Ta có: 1 + $tan^{2}\alpha$ = 1 + $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ = $\frac{cos^{2}\alpha}{cos^{2}\alpha}$ + $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ = $\frac{1}{cos^{2}\alpha}$ (vì $cos^{2}\alpha$ + $sin^{2}\alpha$ = 1 chứng minh câu a)
Vậy 1 + $tan^{2}\alpha$ = $\frac{1}{cos^{2}\alpha}$
d. Ta có: 1 + $cot^{2}\alpha$ = 1 + $\frac{cos^{2}\alpha}{sin^{2}\alpha}$ = $\frac{sin^{2}\alpha}{sin^{2}\alpha}$ + $\frac{cos^{2}\alpha}{sin^{2}\alpha}$ = $\frac{1}{sin^{2}\alpha}$ (vì $cos^{2}\alpha$ + $sin^{2}\alpha$ = 1 chứng minh câu a)
Vậy 1 + $cot^{2}\alpha$ = $\frac{1}{sin^{2}\alpha}$