Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$

Question

find the : $$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$$

My Try : $$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$

$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\circ}}+\dfrac{\cos20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}$$

now what do i do ?

Answer 1

It's $$\dfrac{\sin20^{\circ}}{\sin 20^{\circ}+\cos 20^{\circ}}+\dfrac{\cos20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}=1$$

Answer 2

Let $x=20^{\circ}$

\begin{eqnarray}\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}&=&\dfrac{1}{1+\cot 20^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}\\&=& \dfrac{\sin x}{\sin x+\cos x}+\dfrac{\cos x}{\cos x+\sin x}\\ &=&1\end{eqnarray}

Answer 3

$$\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\tan { { 20 }^{ \circ } } } =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\cot { { 70 }^{ \circ } } } =\\ =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\frac { 1 }{ \tan { { 70 }^{ \circ } } } } =\frac { 1+\tan { { 70 }^{ \circ } } }{ 1+\tan { { 70 }^{ \circ } } } =1$$

Answer 4

See that $$\tan (x) = \frac{\sin x}{\cos x}=\frac{\cos (90^{\circ}-x)}{\sin (90^{\circ}-x)}=\frac{1}{\tan (90^{\circ}-x)}$$ what happen if you take $x=70$

$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}} = \dfrac{1}{1+\frac{1}{\tan 20^{\circ}}}+\dfrac{1}{1+\tan 20^{\circ}}= \dfrac{\tan 20^{\circ}}{1+\tan 20^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}} = 1$$