Question

Prove that $\tan 70^\circ = \cot 70^\circ + 2\cot 40^\circ$.

Answer 1

In this question, we are given a trigonometric equation and we have been asked to prove that LHS = RHS. Start by taking the LHS. Expand the LHS in such a way that you can use the formula $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}$ for expansion. After using the formula, shift the denominator to the other side and multiply it with the LHS. Then simplify the equation using some trigonometric identities. You will get the simplified equation in terms of tan. Convert it into cot using $\tan \left( {90 - x} \right)^\circ = \cot x^\circ$ as our RHS is in terms of cot. This will help in proving LHS = RHS.

Formula used: 1) $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}$
2) $\tan \left( {90 - x} \right)^\circ = \cot x^\circ$

Complete step-by-step answer:
We are given a trigonometric equation and we have been asked to prove that LHS = RHS. We will begin by expanding the LHS of the given equation.
LHS = $\tan 70^\circ$
We can write $\tan 70^\circ = \tan \left( {20 + 50} \right)^\circ$
Now, we will expand this using the formula $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}$.
$\Rightarrow \tan 70^\circ = \dfrac{{\tan 20^\circ + \tan 50^\circ }}{{1 - \tan 20^\circ .\tan 50^\circ }}$
Shifting the denominator to the other side,
$\Rightarrow \tan 70^\circ \left( {1 - \tan 20^\circ .\tan 50^\circ } \right) = \tan 20^\circ + \tan 50^\circ$
Multiplying $\tan 70^\circ$,
$\Rightarrow \tan 70^\circ - \tan 70^\circ \tan 20^\circ \tan 50^\circ = \tan 20^\circ + \tan 50^\circ$
Rearranging terms and keeping only $\tan 70^\circ$ on the LHS,
$\Rightarrow \tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \tan 70^\circ \tan 20^\circ \tan 50^\circ$ ……………….... (1)
We can write $\tan 20^\circ = \tan \left( {90 - 70} \right)^\circ$ ………………... (2)
We know that $\tan \left( {90 - x} \right)^\circ = \cot x^\circ$. Using it in equation (2),
$\Rightarrow \tan \left( {90 - 70} \right)^\circ = \cot 70^\circ$
Therefore, $\tan 20^\circ = \cot 70^\circ$
Now we can put this in equation (1),
$\Rightarrow \tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \tan 70^\circ \cot 70^\circ \tan 50^\circ$
If we expand $\tan 70^\circ \cot 70^\circ$ $= \dfrac{{{\text{sin70}}^\circ \times {\text{cos70}}^\circ }}{{\cos 70^\circ \times \sin 70^\circ }}$
On cancelling we will get, $\dfrac{{{\text{sin70}}^\circ \times {\text{cos70}}^\circ }}{{\cos 70^\circ \times \sin 70^\circ }} = 1$
Therefore, $\tan 70^\circ \cot 70^\circ$ $= 1$. Using this in equation (1) again,
$\Rightarrow \tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \tan 50^\circ$
Simplifying,
$\Rightarrow \tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ$ …………..…. (3)
Now, we will convert the RHS in terms of cot using the same method as used above:
$\Rightarrow \tan 20^\circ = \tan \left( {90 - 70} \right)^\circ = \cot 70^\circ$
$\Rightarrow \tan 50^\circ = \tan \left( {90 - 40} \right)^\circ = \cot 40^\circ$
Substituting them in equation (3),
$\Rightarrow \tan 70^\circ = \cot 70 + 2\cot 40^\circ$= RHS
$\therefore$ LHS = RHS.
Hence proved.

Note: There are certain formulas which will make certain steps easier. They are:
If $a + b = 90^\circ$, then
$\Rightarrow$ $\tan a^\circ = \cot b^\circ$
For example: We know that $30 + 60 = 90$.
$\tan 30^\circ = \cot 60^\circ = \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow \tan a.\tan b = 1$
For example: We know that $30 + 60 = 90$.
$\tan 30^\circ .\tan 60^\circ = ?$
$\dfrac{1}{{\sqrt 3 }} \times \sqrt 3 = 1$
Therefore, $\tan 30^\circ .\tan 60^\circ = 1$
$\Rightarrow \cot a.\cot b = 1$
For example: We know that $30 + 60 = 90$.
$\cot 30^\circ .\cot 60^\circ = ?$
$\dfrac{1}{{\sqrt 3 }} \times \sqrt 3 = 1$
$\therefore \cot 30^\circ .\cot 60^\circ = 1$