Finding $dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$
$begingroup$
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 ?
algebra-precalculus trigonometry
$endgroup$
add a comment |
$begingroup$
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 ?
algebra-precalculus trigonometry
$endgroup$
1
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34
add a comment |
$begingroup$
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 ?
algebra-precalculus trigonometry
$endgroup$
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 ?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 1 at 22:02
greedoid
45.8k1159116
45.8k1159116
asked Jan 24 '18 at 20:22
Fricul38Fricul38
36918
36918
1
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34
add a comment |
1
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34
1
1
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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}\ &=&1end{eqnarray}
$endgroup$
add a comment |
$begingroup$
It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$
$endgroup$
add a comment |
$begingroup$
$$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$
$endgroup$
add a comment |
$begingroup$
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 $x=70$
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}\ &=&1end{eqnarray}
$endgroup$
add a comment |
$begingroup$
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}\ &=&1end{eqnarray}
$endgroup$
add a comment |
$begingroup$
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}\ &=&1end{eqnarray}
$endgroup$
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}\ &=&1end{eqnarray}
answered Jan 24 '18 at 20:26
greedoidgreedoid
45.8k1159116
45.8k1159116
add a comment |
add a comment |
$begingroup$
It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$
$endgroup$
add a comment |
$begingroup$
It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$
$endgroup$
add a comment |
$begingroup$
It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$
$endgroup$
It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$
answered Jan 24 '18 at 20:24
Michael RozenbergMichael Rozenberg
106k1894198
106k1894198
add a comment |
add a comment |
$begingroup$
$$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$
$endgroup$
add a comment |
$begingroup$
$$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$
$endgroup$
add a comment |
$begingroup$
$$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$
$endgroup$
$$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$
answered Jan 24 '18 at 20:29
haqnaturalhaqnatural
20.8k72457
20.8k72457
add a comment |
add a comment |
$begingroup$
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 $x=70$
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$
$endgroup$
add a comment |
$begingroup$
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 $x=70$
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$
$endgroup$
add a comment |
$begingroup$
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 $x=70$
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$
$endgroup$
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 $x=70$
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$
edited Jan 24 '18 at 21:11
answered Jan 24 '18 at 20:40
Guy FsoneGuy Fsone
17.2k43074
17.2k43074
add a comment |
add a comment |
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1
$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24
$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34