On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$ I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.
Thanks!
 A: \begin{align*}
 \tan 70-\tan 20 & =\tan 70-\cot 70\\
 & =\tan 70-\frac{1}{\tan 70}\\
 & = \frac{\tan^2 70-1}{\tan 70}=-\frac{1-\tan^2 70}{\tan 70}\\
 & = -\frac{2(1-\tan^2 70)}{2\tan 70}=-\frac{2}{\frac{2\tan 70}{1-\tan^2 70}}\\
 & = \frac{-2}{\tan 140}= \frac{-2}{-\tan 40}= \frac{2}{\tan 40}=2\cot 40 
 \end{align*}
    Now, $$ \tan 70-\tt-2\tan 40 =2\cot 40-2\tan 40=-2(\tan 40-\cot 40)  $$
    \begin{align*}
 -2(\tan 40-\cot 40)= -2\left( \tan 40-\frac{1}{\tan 40} \right) 
 \end{align*}
    Now do the same as above to get the result. You will get $ 4\cot 80=4\tan 10. $
A: $\tan 70 = \tan(90 - 20) = \cot 20$. So
$$\tan 70 - \tan 20 = \cot 20 - \tan 20=  \frac{\cos^2 20 - \sin^2 20}{\cos 20 \sin 20} = \frac{\cos 40}{\frac12 \sin 40} = 2 \cot 40$$
Repeating the same procedure, we get that the LHS is:
$$2(\cot 40 - \tan 40) = \cdots = 2 (2 \cot 80) = 4\cot (90 - 10) = 4 \tan 10$$
A: $$\begin{align}
\tan  70-\tan  20-2\tan  40 &= \left( \tan { 70-\tan { 40 }  }  \right) -\left( \tan { 40+\tan { 20 }  }  \right) \\ 
&=\frac { \sin { 70 }  }{ \cos { 70 }  } -\frac { \sin { 40 }  }{ \cos { 40 }  } -\left( \frac { \sin { 40 }  }{ \cos { 40 }  } +\frac { \sin { 20 }  }{ \cos { 20 }  }  \right) \\
&=\frac { \sin { 30 }  }{ \cos { 70\cos { 40 }  }  }  -\frac { \sin { 60 }  }{ \cos { 20\cos { 40 }  }  } \\
&=\frac { 1 }{ 2\cos { 40 }  } \left( \frac { 1 }{ \cos { 70 }  } -\frac { \sqrt { 3 }  }{ \cos { 20 }  }  \right) \\ 
&=\frac { 1 }{ 2\cos { 40 }  } \left( \frac { \cos { 20-\sqrt { 3 } \cos { 70 }  }  }{ \cos { 70\cos { 20 }  }  }  \right) \\
&=\frac { 1 }{ 2\cos { 40 }  } \left( \frac { \cos { 20-\sqrt { 3 } \sin { 20 }  }  }{ \sin { 20 } \cos { 20 }  }  \right) \\ 
&=\frac { 1 }{ \cos { 40 }  } \left( \frac { \frac { 1 }{ 2 } \cos { 20 } -\frac { \sqrt { 3 }  }{ 2 } \sin { 20 }  }{ \sin { 20 } \cos { 20 }  }  \right) \\
&=\frac { 2 }{ \cos { 40 }  } \frac { \cos { 60\cos { 20 } -\sin { 60\sin { 20 }  }  }  }{ 2\sin { 20 } \cos { 20 }  } \\ 
&=\frac { 2 }{ \cos { 40 }  } \frac { \cos { 80 }  }{ \sin { 40 }  } \\
&= 4\frac { \cos { 80 }  }{ 2\sin { 40 } \cos { 40 }  } \\
&= 4\frac { \cos { 80 }  }{ \sin { 80 }  } =4\cot { 80 } =4\tan { 10 } 
\end{align}$$
A: 
Note:- $\tan A\cdot\tan B=1$, $A+B=90°$

Now, as $70°+20°=90°$, $\therefore \tan 70°\tan20°=1$, so $\tan 70°=\dfrac{1}{\tan20°}$
$$\begin{aligned}
\tan 70 ° -\tan 20° -2 \tan 40°
&=\dfrac{1}{\tan 20°} -\tan 20° -2 \tan 40° \\
&=\dfrac{2(1-\tan^220°)}{2\tan 20°} -2 \tan 40° \\
&=2 \cot 40° -2 \tan 40°=2(\dfrac{1-\tan^2{40°}}{\tan 40°}) \\
&=4(\dfrac{1-\tan^2{40°}}{2\tan 40°})=4\cot{80°}=4\tan{(90°-80°)} \\
&=\boxed{4\tan{10°}}=RHS\qquad \text{Hence, proved}
\end{aligned}$$
A: $$\cot A-\tan A=\dfrac{\cos^2A-\sin^2A}{\sin A\cos A}=2\cot2A$$
$\tan70^\circ=\cot20^\circ, A=20^\circ\implies ?$
Again set $A=40^\circ$
Finally $\cot80^\circ=\tan(90-80)^\circ$
A: As we know there is a formula
$$\tan A-\tan B=2\tan{(A-B)} $$
If $A+B=90$
So according to question,
$$2\tan 50-2\tan 40=4\tan 10$$
$$2(tan 50-tan 40)=4tan 10$$
$$2.2tan 10=4tan 10$$
$$4tan 10=4tan 10$$
$$1=1$$
