Question

$\frac{\sin {70}^0+\cos {40}^0}{\cos {70}^0+\sin {40}^0}=$

Answer 1

$\begin{array}{cc}\frac{\sin 70+\cos 40}{\cos 70+\sin 40}=\frac{\sin 70+\sin 50}{\cos 70+\cos 50}& \\ =\frac{\sin \left(60+10\right)+\sin \left(60-10\right)}{\cos \left(60+10\right)+\cos \left(60-10\right)}& \left\{\begin{array}{c}u\sin gformulae\\ \sin \left(\frac{\pi }{2}-\theta \right)=\cos x\end{array}\right\}\\ =\frac{2\sin 60\cos 10}{2.\cos 60\cos 10}& \\ =\tan 60& \\ =\sqrt{3}.& \end{array}$