Question

4. Найдите значение выражения: tg15⁰ + ctg15⁰

Answer 1

Основные формулы:

$\sin^2x+\cos^2x=1$

$\cos(-x)=\cos x$

$\mathrm{ctg}{\,}(90^\circ-x) =\mathrm{tg}{\,}x$

$\mathrm{tg}{\,}x +\mathrm{tg}{\,}y=\dfrac{\sin(x+y)}{\cos x\cos y}$

$\cos x\cos y=\dfrac{1}{2} \Big( \cos(x+y)+\cos(x-y)\Big)$

Первый способ:

$\mathrm{tg}{\,}15^\circ +\mathrm{ctg}{\,}15^\circ =\dfrac{\sin15^\circ}{\cos15^\circ} +\dfrac{\cos15^\circ}{\sin15^\circ} =\dfrac{\sin^215^\circ+\cos^215^\circ}{\sin15^\circ\cos15^\circ} =$

$=\dfrac{1}{\sin15^\circ\cos15^\circ} =\dfrac{2}{2\sin15^\circ\cos15^\circ} =\dfrac{2}{\sin30^\circ} =\dfrac{2}{1/2} =\boxed{4}$

Второй способ:

$\mathrm{tg}{\,}15^\circ +\mathrm{ctg}{\,}15^\circ =\mathrm{tg}{\,}15^\circ +\mathrm{ctg}{\,}(90^\circ-75^\circ) =\mathrm{tg}{\,}15^\circ +\mathrm{tg}{\,}75^\circ =$

$=\dfrac{\sin(15^\circ +75^\circ) }{\cos15^\circ\cos 75^\circ} =\dfrac{\sin90^\circ }{\dfrac{1}{2}\cdot\big(\cos(15^\circ+75^\circ)+\cos(15^\circ-75^\circ)\big)} =$

$=\dfrac{1 }{\dfrac{1}{2}\cdot\big(\cos90^\circ+\cos(-60^\circ)\big)} =\dfrac{1 }{\dfrac{1}{2}\cdot\big(0+\cos60^\circ\big)} =\dfrac{1 }{\dfrac{1}{2}\cdot\left(0+\dfrac{1}{2} \right)} =\boxed{4}$

Ответ: 4

Answer 2

Ответ: 4

Объяснение:

$tg15+ ctg15= \frac{sin15}{cos15}+\frac{cos 15}{sin 15}= \frac{sin^215+cos^215}{cos15*sin15} =\frac{1}{0.5sin30} = \frac{1}{0.5*0.5}=4$