On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$

Question

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!

Answer 1

\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. $

Answer 2

$\tan 70 = \tan(90 - 20) = \cot 20$. 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:

$$2(\cot 40 - \tan 40) = \cdots = 2 (2 \cot 80) = 4\cot (90 - 10) = 4 \tan 10$$

Answer 3

$$\begin{align} \tan 70-\tan 20-2\tan 40 &= \left( \tan { 70-\tan { 40 } } \right) -\left( \tan { 40+\tan { 20 } } \right) \\ &=\frac { \sin { 70 } }{ \cos { 70 } } -\frac { \sin { 40 } }{ \cos { 40 } } -\left( \frac { \sin { 40 } }{ \cos { 40 } } +\frac { \sin { 20 } }{ \cos { 20 } } \right) \\ &=\frac { \sin { 30 } }{ \cos { 70\cos { 40 } } } -\frac { \sin { 60 } }{ \cos { 20\cos { 40 } } } \\ &=\frac { 1 }{ 2\cos { 40 } } \left( \frac { 1 }{ \cos { 70 } } -\frac { \sqrt { 3 } }{ \cos { 20 } } \right) \\ &=\frac { 1 }{ 2\cos { 40 } } \left( \frac { \cos { 20-\sqrt { 3 } \cos { 70 } } }{ \cos { 70\cos { 20 } } } \right) \\ &=\frac { 1 }{ 2\cos { 40 } } \left( \frac { \cos { 20-\sqrt { 3 } \sin { 20 } } }{ \sin { 20 } \cos { 20 } } \right) \\ &=\frac { 1 }{ \cos { 40 } } \left( \frac { \frac { 1 }{ 2 } \cos { 20 } -\frac { \sqrt { 3 } }{ 2 } \sin { 20 } }{ \sin { 20 } \cos { 20 } } \right) \\ &=\frac { 2 }{ \cos { 40 } } \frac { \cos { 60\cos { 20 } -\sin { 60\sin { 20 } } } }{ 2\sin { 20 } \cos { 20 } } \\ &=\frac { 2 }{ \cos { 40 } } \frac { \cos { 80 } }{ \sin { 40 } } \\ &= 4\frac { \cos { 80 } }{ 2\sin { 40 } \cos { 40 } } \\ &= 4\frac { \cos { 80 } }{ \sin { 80 } } =4\cot { 80 } =4\tan { 10 } \end{align}$$

Answer 4

Note:- $\tan A\cdot\tan B=1$, $A+B=90°$

Now, as $70°+20°=90°$, $\therefore \tan 70°\tan20°=1$, so $\tan 70°=\dfrac{1}{\tan20°}$

$$\begin{aligned} \tan 70 ° -\tan 20° -2 \tan 40° &=\dfrac{1}{\tan 20°} -\tan 20° -2 \tan 40° \\ &=\dfrac{2(1-\tan^220°)}{2\tan 20°} -2 \tan 40° \\ &=2 \cot 40° -2 \tan 40°=2(\dfrac{1-\tan^2{40°}}{\tan 40°}) \\ &=4(\dfrac{1-\tan^2{40°}}{2\tan 40°})=4\cot{80°}=4\tan{(90°-80°)} \\ &=\boxed{4\tan{10°}}=RHS\qquad \text{Hence, proved} \end{aligned}$$

Answer 5

$$\cot A-\tan A=\dfrac{\cos^2A-\sin^2A}{\sin A\cos A}=2\cot2A$$

$\tan70^\circ=\cot20^\circ, A=20^\circ\implies ?$

Again set $A=40^\circ$

Finally $\cot80^\circ=\tan(90-80)^\circ$

Answer 6

As we know there is a formula $$\tan A-\tan B=2\tan{(A-B)} $$ If $A+B=90$

So according to question, $$2\tan 50-2\tan 40=4\tan 10$$ $$2(tan 50-tan 40)=4tan 10$$ $$2.2tan 10=4tan 10$$ $$4tan 10=4tan 10$$ $$1=1$$