Ответы
Ответ:
Если угол α<90°:
\displaystyle tg\alpha=\frac{\sin\alpha}{\cos\alpha} =\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha} } =\frac{4}{5\sqrt{1-\frac{16}{25} } } =\frac{4}{5\sqrt{\frac{9}{25} } }=\frac{4}{5\cdot\frac{3}{5} } = \frac{4\cdot5}{3\cdot5} =\frac{4}{3}tgα=
cosα
sinα
=
1−sin
2
α
sinα
=
5
1−
25
16
4
=
5
25
9
4
=
5⋅
5
3
4
=
3⋅5
4⋅5
=
3
4
2. Если угол α>90°:
\displaystyle tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sin(180^\circ-\alpha)}{\cos(180^\circ-\alpha)}=\frac{\sin\alpha}{-\cos\alpha}=\frac{\sin\alpha}{-\sqrt{1-\sin^2\alpha}}=-\frac{4}{5\cdot\sqrt{1-\frac{16}{25}}} =-\frac{4}{5\cdot\sqrt{\frac{9}{25}}}=-\frac{4}{5\cdot\frac{3}{5}} =-\frac{4\cdot5}{3\cdot5}=-\frac{4}{3}tgα=
cosα
sinα
=
cos(180
∘
−α)
sin(180
∘
−α)
=
−cosα
sinα
=
−
1−sin
2
α
sinα
=−
5⋅
1−
25
16
4
=−
5⋅
25
9
4
=−
5⋅
5
3
4
=−
3⋅5
4⋅5
=−
3
4