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show me the process of solving (1-cot200)(1-cot25)

Guest Jul 1, 2015

Best Answer 

 #1
avatar+18712 
+16

show me the process of solving (1-cot200)(1-cot25)

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]\\\\
&=& \left[ 1 -
\dfrac { 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[ 1 -
\dfrac { 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=& \left[
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[
\dfrac { \tan{ ( 25\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=&
\dfrac { \left[\tan{ ( 200\ensurement{^{\circ}} ) } - 1\right]\cdot \left[ \tan{ ( 25\ensurement{^{\circ}} ) } - 1 \right] }
{ \tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \tan{ ( 200\ensurement{^{\circ}} ) } - \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

 

$$\\\small{ \text{Formula: $
\begin{array}{rcl}
&&\\
&&\\
&&\\
&&\\
\boxed{
\tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
=
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }
\qquad \tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
= \tan{ (225\ensurement{^{\circ}} ) }
= \tan{ (180\ensurement{^{\circ}}+45\ensurement{^{\circ}} ) }
= \tan{ (45\ensurement{^{\circ}} ) }
= 1
}
\end{array}
$}} \\\\
\small{ \text{$
\begin{array}{rcl}
1 &=&
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
&=&
\tan{ ( 200\ensurement{^{\circ}} ) + \tan{ ( 25\ensurement{^{\circ}} ) } } \\\\
\end{array}
$}}$$

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
\right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
-1 + \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+ \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
2\cdot \dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\mathbf{
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
}&\mathbf{
=}& \mathbf{2 }
\end{array}
$}}$$

 

heureka  Jul 2, 2015
Sort: 

2+0 Answers

 #1
avatar+18712 
+16
Best Answer

show me the process of solving (1-cot200)(1-cot25)

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]\\\\
&=& \left[ 1 -
\dfrac { 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[ 1 -
\dfrac { 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=& \left[
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[
\dfrac { \tan{ ( 25\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=&
\dfrac { \left[\tan{ ( 200\ensurement{^{\circ}} ) } - 1\right]\cdot \left[ \tan{ ( 25\ensurement{^{\circ}} ) } - 1 \right] }
{ \tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \tan{ ( 200\ensurement{^{\circ}} ) } - \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

 

$$\\\small{ \text{Formula: $
\begin{array}{rcl}
&&\\
&&\\
&&\\
&&\\
\boxed{
\tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
=
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }
\qquad \tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
= \tan{ (225\ensurement{^{\circ}} ) }
= \tan{ (180\ensurement{^{\circ}}+45\ensurement{^{\circ}} ) }
= \tan{ (45\ensurement{^{\circ}} ) }
= 1
}
\end{array}
$}} \\\\
\small{ \text{$
\begin{array}{rcl}
1 &=&
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
&=&
\tan{ ( 200\ensurement{^{\circ}} ) + \tan{ ( 25\ensurement{^{\circ}} ) } } \\\\
\end{array}
$}}$$

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
\right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
-1 + \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

 

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+ \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
2\cdot \dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\mathbf{
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
}&\mathbf{
=}& \mathbf{2 }
\end{array}
$}}$$

 

heureka  Jul 2, 2015
 #2
avatar+90988 
0

That is great Heureka    

Melody  Jul 2, 2015

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