Question

como verifico se a identidade 1+tg a =$\frac{ \sqrt{2}sen(a-45°)}{cos a}$

Answer 1

Identidades utilizadas:

$\begin{array}{rclc} \cos \alpha&=&\mathrm{sen}\left(\alpha+90^{\circ} \right )&\;\;\;\;\;\;\;\text{(i)}\\ \\ \mathrm{sen\,}\alpha + \mathrm{sen\,} \beta&=&2\cdot \mathrm{sen}\left(\dfrac{\alpha + \beta}{2} \right )\cdot \cos \left(\dfrac{\alpha-\beta}{2} \right )&\;\;\;\;\;\;\;\text{(ii)} \end{array}$


$1+\mathrm{tg\,}a\\ \\ =1+\dfrac{\mathrm{sen\,}a}{\cos a}\\ \\ =\dfrac{\cos a + \mathrm{sen\,}a}{\cos a}\\ \\ =\dfrac{\mathrm{sen}\left(a+90^{\circ} \right ) + \mathrm{sen\,}a}{\cos a}\\ \\ =\dfrac{2\cdot \mathrm{sen}\left(\frac{\left(a+90^{\circ} \right )+a}{2} \right )\cdot \cos \left(\frac{\left(a+90^{\circ} \right )-a}{2} \right )}{\cos a}\\ \\ =\dfrac{2\cdot \mathrm{sen}\left(\frac{2a+90^{\circ}}{2} \right )\cdot \cos \left(\frac{90^{\circ}}{2} \right )}{\cos a}\\ \\ =\dfrac{2\cdot \mathrm{sen}\left(\frac{\not 2\left(a+45^{\circ} \right )}{\not 2} \right )\cdot \cos 45^{\circ}}{\cos a}$

$=\dfrac{2\cdot \mathrm{sen}\left(a+45^{\circ} \right )\cdot \frac{\sqrt{2}}{2}}{\cos a}\\ \\ =\dfrac{\not 2 \cdot \frac{\sqrt{2}}{\not 2}\cdot \mathrm{sen}\left(a+45^{\circ} \right )}{\cos a}\\ \\ =\dfrac{\sqrt{2}\mathrm{\,sen}\left(a+45^{\circ} \right )}{\cos a}\\ \\ \\ \boxed{1+\mathrm{tg\,}a=\dfrac{\sqrt{2}\mathrm{\,sen}\left(a+45^{\circ} \right )}{\cos a}}$